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1 | 2818-2821 | Note that the function f is
increasing (i e , f β²(x) > 0) in the interval (c β h, c) and decreasing (i e |
1 | 2819-2822 | e , f β²(x) > 0) in the interval (c β h, c) and decreasing (i e , f β²(x) < 0) in the
interval (c, c + h) |
1 | 2820-2823 | , f β²(x) > 0) in the interval (c β h, c) and decreasing (i e , f β²(x) < 0) in the
interval (c, c + h) This suggests that f β²(c) must be zero |
1 | 2821-2824 | e , f β²(x) < 0) in the
interval (c, c + h) This suggests that f β²(c) must be zero Fig 6 |
1 | 2822-2825 | , f β²(x) < 0) in the
interval (c, c + h) This suggests that f β²(c) must be zero Fig 6 11
Fig 6 |
1 | 2823-2826 | This suggests that f β²(c) must be zero Fig 6 11
Fig 6 12
Rationalised 2023-24
MATHEMATICS
164
Fig 6 |
1 | 2824-2827 | Fig 6 11
Fig 6 12
Rationalised 2023-24
MATHEMATICS
164
Fig 6 13
Similarly, if c is a point of local minima of f , then the graph of f around c will be as
shown in Fig 6 |
1 | 2825-2828 | 11
Fig 6 12
Rationalised 2023-24
MATHEMATICS
164
Fig 6 13
Similarly, if c is a point of local minima of f , then the graph of f around c will be as
shown in Fig 6 14(b) |
1 | 2826-2829 | 12
Rationalised 2023-24
MATHEMATICS
164
Fig 6 13
Similarly, if c is a point of local minima of f , then the graph of f around c will be as
shown in Fig 6 14(b) Here f is decreasing (i |
1 | 2827-2830 | 13
Similarly, if c is a point of local minima of f , then the graph of f around c will be as
shown in Fig 6 14(b) Here f is decreasing (i e |
1 | 2828-2831 | 14(b) Here f is decreasing (i e , f β²(x) < 0) in the interval (c β h, c) and
increasing (i |
1 | 2829-2832 | Here f is decreasing (i e , f β²(x) < 0) in the interval (c β h, c) and
increasing (i e |
1 | 2830-2833 | e , f β²(x) < 0) in the interval (c β h, c) and
increasing (i e , f β²(x) > 0) in the interval (c, c + h) |
1 | 2831-2834 | , f β²(x) < 0) in the interval (c β h, c) and
increasing (i e , f β²(x) > 0) in the interval (c, c + h) This again suggest that f β²(c) must
be zero |
1 | 2832-2835 | e , f β²(x) > 0) in the interval (c, c + h) This again suggest that f β²(c) must
be zero The above discussion lead us to the following theorem (without proof) |
1 | 2833-2836 | , f β²(x) > 0) in the interval (c, c + h) This again suggest that f β²(c) must
be zero The above discussion lead us to the following theorem (without proof) Theorem 2 Let f be a function defined on an open interval I |
1 | 2834-2837 | This again suggest that f β²(c) must
be zero The above discussion lead us to the following theorem (without proof) Theorem 2 Let f be a function defined on an open interval I Suppose c β I be any
point |
1 | 2835-2838 | The above discussion lead us to the following theorem (without proof) Theorem 2 Let f be a function defined on an open interval I Suppose c β I be any
point If f has a local maxima or a local minima at x = c, then either f β²(c) = 0 or f is not
differentiable at c |
1 | 2836-2839 | Theorem 2 Let f be a function defined on an open interval I Suppose c β I be any
point If f has a local maxima or a local minima at x = c, then either f β²(c) = 0 or f is not
differentiable at c Remark The converse of above theorem need not
be true, that is, a point at which the derivative vanishes
need not be a point of local maxima or local minima |
1 | 2837-2840 | Suppose c β I be any
point If f has a local maxima or a local minima at x = c, then either f β²(c) = 0 or f is not
differentiable at c Remark The converse of above theorem need not
be true, that is, a point at which the derivative vanishes
need not be a point of local maxima or local minima For example, if f (x) = x3, then f β²(x) = 3x2 and so
f β²(0) = 0 |
1 | 2838-2841 | If f has a local maxima or a local minima at x = c, then either f β²(c) = 0 or f is not
differentiable at c Remark The converse of above theorem need not
be true, that is, a point at which the derivative vanishes
need not be a point of local maxima or local minima For example, if f (x) = x3, then f β²(x) = 3x2 and so
f β²(0) = 0 But 0 is neither a point of local maxima nor
a point of local minima (Fig 6 |
1 | 2839-2842 | Remark The converse of above theorem need not
be true, that is, a point at which the derivative vanishes
need not be a point of local maxima or local minima For example, if f (x) = x3, then f β²(x) = 3x2 and so
f β²(0) = 0 But 0 is neither a point of local maxima nor
a point of local minima (Fig 6 13) |
1 | 2840-2843 | For example, if f (x) = x3, then f β²(x) = 3x2 and so
f β²(0) = 0 But 0 is neither a point of local maxima nor
a point of local minima (Fig 6 13) ANote A point c in the domain of a function f at
which either f β²(c) = 0 or f is not differentiable is
called a critical point of f |
1 | 2841-2844 | But 0 is neither a point of local maxima nor
a point of local minima (Fig 6 13) ANote A point c in the domain of a function f at
which either f β²(c) = 0 or f is not differentiable is
called a critical point of f Note that if f is continuous
at c and f β²(c) = 0, then there exists an h > 0 such
that f is differentiable in the interval
(c β h, c + h) |
1 | 2842-2845 | 13) ANote A point c in the domain of a function f at
which either f β²(c) = 0 or f is not differentiable is
called a critical point of f Note that if f is continuous
at c and f β²(c) = 0, then there exists an h > 0 such
that f is differentiable in the interval
(c β h, c + h) We shall now give a working rule for finding points of local maxima or points of
local minima using only the first order derivatives |
1 | 2843-2846 | ANote A point c in the domain of a function f at
which either f β²(c) = 0 or f is not differentiable is
called a critical point of f Note that if f is continuous
at c and f β²(c) = 0, then there exists an h > 0 such
that f is differentiable in the interval
(c β h, c + h) We shall now give a working rule for finding points of local maxima or points of
local minima using only the first order derivatives Theorem 3 (First Derivative Test) Let f be a function defined on an open interval I |
1 | 2844-2847 | Note that if f is continuous
at c and f β²(c) = 0, then there exists an h > 0 such
that f is differentiable in the interval
(c β h, c + h) We shall now give a working rule for finding points of local maxima or points of
local minima using only the first order derivatives Theorem 3 (First Derivative Test) Let f be a function defined on an open interval I Let f be continuous at a critical point c in I |
1 | 2845-2848 | We shall now give a working rule for finding points of local maxima or points of
local minima using only the first order derivatives Theorem 3 (First Derivative Test) Let f be a function defined on an open interval I Let f be continuous at a critical point c in I Then
(i)
If f β²(x) changes sign from positive to negative as x increases through c, i |
1 | 2846-2849 | Theorem 3 (First Derivative Test) Let f be a function defined on an open interval I Let f be continuous at a critical point c in I Then
(i)
If f β²(x) changes sign from positive to negative as x increases through c, i e |
1 | 2847-2850 | Let f be continuous at a critical point c in I Then
(i)
If f β²(x) changes sign from positive to negative as x increases through c, i e , if
f β²(x) > 0 at every point sufficiently close to and to the left of c, and f β²(x) < 0 at
every point sufficiently close to and to the right of c, then c is a point of local
maxima |
1 | 2848-2851 | Then
(i)
If f β²(x) changes sign from positive to negative as x increases through c, i e , if
f β²(x) > 0 at every point sufficiently close to and to the left of c, and f β²(x) < 0 at
every point sufficiently close to and to the right of c, then c is a point of local
maxima (ii)
If f β²(x) changes sign from negative to positive as x increases through c, i |
1 | 2849-2852 | e , if
f β²(x) > 0 at every point sufficiently close to and to the left of c, and f β²(x) < 0 at
every point sufficiently close to and to the right of c, then c is a point of local
maxima (ii)
If f β²(x) changes sign from negative to positive as x increases through c, i e |
1 | 2850-2853 | , if
f β²(x) > 0 at every point sufficiently close to and to the left of c, and f β²(x) < 0 at
every point sufficiently close to and to the right of c, then c is a point of local
maxima (ii)
If f β²(x) changes sign from negative to positive as x increases through c, i e , if
f β²(x) < 0 at every point sufficiently close to and to the left of c, and f β²(x) > 0 at
every point sufficiently close to and to the right of c, then c is a point of local
minima |
1 | 2851-2854 | (ii)
If f β²(x) changes sign from negative to positive as x increases through c, i e , if
f β²(x) < 0 at every point sufficiently close to and to the left of c, and f β²(x) > 0 at
every point sufficiently close to and to the right of c, then c is a point of local
minima (iii)
If f β²(x) does not change sign as x increases through c, then c is neither a point of
local maxima nor a point of local minima |
1 | 2852-2855 | e , if
f β²(x) < 0 at every point sufficiently close to and to the left of c, and f β²(x) > 0 at
every point sufficiently close to and to the right of c, then c is a point of local
minima (iii)
If f β²(x) does not change sign as x increases through c, then c is neither a point of
local maxima nor a point of local minima Infact, such a point is called point of
inflection (Fig 6 |
1 | 2853-2856 | , if
f β²(x) < 0 at every point sufficiently close to and to the left of c, and f β²(x) > 0 at
every point sufficiently close to and to the right of c, then c is a point of local
minima (iii)
If f β²(x) does not change sign as x increases through c, then c is neither a point of
local maxima nor a point of local minima Infact, such a point is called point of
inflection (Fig 6 13) |
1 | 2854-2857 | (iii)
If f β²(x) does not change sign as x increases through c, then c is neither a point of
local maxima nor a point of local minima Infact, such a point is called point of
inflection (Fig 6 13) Rationalised 2023-24
APPLICATION OF DERIVATIVES
165
ANote If c is a point of local maxima of f , then f (c) is a local maximum value of
f |
1 | 2855-2858 | Infact, such a point is called point of
inflection (Fig 6 13) Rationalised 2023-24
APPLICATION OF DERIVATIVES
165
ANote If c is a point of local maxima of f , then f (c) is a local maximum value of
f Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f |
1 | 2856-2859 | 13) Rationalised 2023-24
APPLICATION OF DERIVATIVES
165
ANote If c is a point of local maxima of f , then f (c) is a local maximum value of
f Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f Figures 6 |
1 | 2857-2860 | Rationalised 2023-24
APPLICATION OF DERIVATIVES
165
ANote If c is a point of local maxima of f , then f (c) is a local maximum value of
f Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f Figures 6 13 and 6 |
1 | 2858-2861 | Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f Figures 6 13 and 6 14, geometrically explain Theorem 3 |
1 | 2859-2862 | Figures 6 13 and 6 14, geometrically explain Theorem 3 Fig 6 |
1 | 2860-2863 | 13 and 6 14, geometrically explain Theorem 3 Fig 6 14
Example 17 Find all points of local maxima and local minima of the function f
given by
f (x) = x3 β 3x + 3 |
1 | 2861-2864 | 14, geometrically explain Theorem 3 Fig 6 14
Example 17 Find all points of local maxima and local minima of the function f
given by
f (x) = x3 β 3x + 3 Solution We have
f (x) = x3 β 3x + 3
or
f β²(x) = 3x2 β 3 = 3(x β 1) (x + 1)
or
f β²(x) = 0 at x = 1 and x = β 1
Thus, x = Β± 1 are the only critical points which could possibly be the points of local
maxima and/or local minima of f |
1 | 2862-2865 | Fig 6 14
Example 17 Find all points of local maxima and local minima of the function f
given by
f (x) = x3 β 3x + 3 Solution We have
f (x) = x3 β 3x + 3
or
f β²(x) = 3x2 β 3 = 3(x β 1) (x + 1)
or
f β²(x) = 0 at x = 1 and x = β 1
Thus, x = Β± 1 are the only critical points which could possibly be the points of local
maxima and/or local minima of f Let us first examine the point x = 1 |
1 | 2863-2866 | 14
Example 17 Find all points of local maxima and local minima of the function f
given by
f (x) = x3 β 3x + 3 Solution We have
f (x) = x3 β 3x + 3
or
f β²(x) = 3x2 β 3 = 3(x β 1) (x + 1)
or
f β²(x) = 0 at x = 1 and x = β 1
Thus, x = Β± 1 are the only critical points which could possibly be the points of local
maxima and/or local minima of f Let us first examine the point x = 1 Note that for values close to 1 and to the right of 1, f β²(x) > 0 and for values close
to 1 and to the left of 1, f β²(x) < 0 |
1 | 2864-2867 | Solution We have
f (x) = x3 β 3x + 3
or
f β²(x) = 3x2 β 3 = 3(x β 1) (x + 1)
or
f β²(x) = 0 at x = 1 and x = β 1
Thus, x = Β± 1 are the only critical points which could possibly be the points of local
maxima and/or local minima of f Let us first examine the point x = 1 Note that for values close to 1 and to the right of 1, f β²(x) > 0 and for values close
to 1 and to the left of 1, f β²(x) < 0 Therefore, by first derivative test, x = 1 is a point
of local minima and local minimum value is f (1) = 1 |
1 | 2865-2868 | Let us first examine the point x = 1 Note that for values close to 1 and to the right of 1, f β²(x) > 0 and for values close
to 1 and to the left of 1, f β²(x) < 0 Therefore, by first derivative test, x = 1 is a point
of local minima and local minimum value is f (1) = 1 In the case of x = β1, note that
f β²(x) > 0, for values close to and to the left of β1 and f β²(x) < 0, for values close to and
to the right of β 1 |
1 | 2866-2869 | Note that for values close to 1 and to the right of 1, f β²(x) > 0 and for values close
to 1 and to the left of 1, f β²(x) < 0 Therefore, by first derivative test, x = 1 is a point
of local minima and local minimum value is f (1) = 1 In the case of x = β1, note that
f β²(x) > 0, for values close to and to the left of β1 and f β²(x) < 0, for values close to and
to the right of β 1 Therefore, by first derivative test, x = β 1 is a point of local maxima
and local maximum value is f (β1) = 5 |
1 | 2867-2870 | Therefore, by first derivative test, x = 1 is a point
of local minima and local minimum value is f (1) = 1 In the case of x = β1, note that
f β²(x) > 0, for values close to and to the left of β1 and f β²(x) < 0, for values close to and
to the right of β 1 Therefore, by first derivative test, x = β 1 is a point of local maxima
and local maximum value is f (β1) = 5 Values of x
Sign of f β²β²β²β²β²(x) = 3(x β 1) (x + 1)
Close to 1
to the right (say 1 |
1 | 2868-2871 | In the case of x = β1, note that
f β²(x) > 0, for values close to and to the left of β1 and f β²(x) < 0, for values close to and
to the right of β 1 Therefore, by first derivative test, x = β 1 is a point of local maxima
and local maximum value is f (β1) = 5 Values of x
Sign of f β²β²β²β²β²(x) = 3(x β 1) (x + 1)
Close to 1
to the right (say 1 1 etc |
1 | 2869-2872 | Therefore, by first derivative test, x = β 1 is a point of local maxima
and local maximum value is f (β1) = 5 Values of x
Sign of f β²β²β²β²β²(x) = 3(x β 1) (x + 1)
Close to 1
to the right (say 1 1 etc )
>0
to the left (say 0 |
1 | 2870-2873 | Values of x
Sign of f β²β²β²β²β²(x) = 3(x β 1) (x + 1)
Close to 1
to the right (say 1 1 etc )
>0
to the left (say 0 9 etc |
1 | 2871-2874 | 1 etc )
>0
to the left (say 0 9 etc )
<0
Close to β1
to the right (say
0 |
1 | 2872-2875 | )
>0
to the left (say 0 9 etc )
<0
Close to β1
to the right (say
0 9 etc |
1 | 2873-2876 | 9 etc )
<0
Close to β1
to the right (say
0 9 etc )
0
to the left (say
1 |
1 | 2874-2877 | )
<0
Close to β1
to the right (say
0 9 etc )
0
to the left (say
1 1 etc |
1 | 2875-2878 | 9 etc )
0
to the left (say
1 1 etc )
0
β
<
β
>
Rationalised 2023-24
MATHEMATICS
166
Fig 6 |
1 | 2876-2879 | )
0
to the left (say
1 1 etc )
0
β
<
β
>
Rationalised 2023-24
MATHEMATICS
166
Fig 6 15
Example 18 Find all the points of local maxima and local minima of the function f
given by
f (x) = 2x3 β 6x2 + 6x +5 |
1 | 2877-2880 | 1 etc )
0
β
<
β
>
Rationalised 2023-24
MATHEMATICS
166
Fig 6 15
Example 18 Find all the points of local maxima and local minima of the function f
given by
f (x) = 2x3 β 6x2 + 6x +5 Solution We have
f (x) = 2x3 β 6x2 + 6x + 5
or
f β²(x) = 6x2 β 12x + 6 = 6(x β 1)2
or
f β²(x) = 0 at x = 1
Thus, x = 1 is the only critical point of f |
1 | 2878-2881 | )
0
β
<
β
>
Rationalised 2023-24
MATHEMATICS
166
Fig 6 15
Example 18 Find all the points of local maxima and local minima of the function f
given by
f (x) = 2x3 β 6x2 + 6x +5 Solution We have
f (x) = 2x3 β 6x2 + 6x + 5
or
f β²(x) = 6x2 β 12x + 6 = 6(x β 1)2
or
f β²(x) = 0 at x = 1
Thus, x = 1 is the only critical point of f We shall now examine this point for local
maxima and/or local minima of f |
1 | 2879-2882 | 15
Example 18 Find all the points of local maxima and local minima of the function f
given by
f (x) = 2x3 β 6x2 + 6x +5 Solution We have
f (x) = 2x3 β 6x2 + 6x + 5
or
f β²(x) = 6x2 β 12x + 6 = 6(x β 1)2
or
f β²(x) = 0 at x = 1
Thus, x = 1 is the only critical point of f We shall now examine this point for local
maxima and/or local minima of f Observe that f β²(x) β₯ 0, for all x β R and in particular
f β²(x) > 0, for values close to 1 and to the left and to the right of 1 |
1 | 2880-2883 | Solution We have
f (x) = 2x3 β 6x2 + 6x + 5
or
f β²(x) = 6x2 β 12x + 6 = 6(x β 1)2
or
f β²(x) = 0 at x = 1
Thus, x = 1 is the only critical point of f We shall now examine this point for local
maxima and/or local minima of f Observe that f β²(x) β₯ 0, for all x β R and in particular
f β²(x) > 0, for values close to 1 and to the left and to the right of 1 Therefore, by first
derivative test, the point x = 1 is neither a point of local maxima nor a point of local
minima |
1 | 2881-2884 | We shall now examine this point for local
maxima and/or local minima of f Observe that f β²(x) β₯ 0, for all x β R and in particular
f β²(x) > 0, for values close to 1 and to the left and to the right of 1 Therefore, by first
derivative test, the point x = 1 is neither a point of local maxima nor a point of local
minima Hence x = 1 is a point of inflexion |
1 | 2882-2885 | Observe that f β²(x) β₯ 0, for all x β R and in particular
f β²(x) > 0, for values close to 1 and to the left and to the right of 1 Therefore, by first
derivative test, the point x = 1 is neither a point of local maxima nor a point of local
minima Hence x = 1 is a point of inflexion Remark One may note that since f β²(x), in Example 30, never changes its sign on R,
graph of f has no turning points and hence no point of local maxima or local minima |
1 | 2883-2886 | Therefore, by first
derivative test, the point x = 1 is neither a point of local maxima nor a point of local
minima Hence x = 1 is a point of inflexion Remark One may note that since f β²(x), in Example 30, never changes its sign on R,
graph of f has no turning points and hence no point of local maxima or local minima We shall now give another test to examine local maxima and local minima of a
given function |
1 | 2884-2887 | Hence x = 1 is a point of inflexion Remark One may note that since f β²(x), in Example 30, never changes its sign on R,
graph of f has no turning points and hence no point of local maxima or local minima We shall now give another test to examine local maxima and local minima of a
given function This test is often easier to apply than the first derivative test |
1 | 2885-2888 | Remark One may note that since f β²(x), in Example 30, never changes its sign on R,
graph of f has no turning points and hence no point of local maxima or local minima We shall now give another test to examine local maxima and local minima of a
given function This test is often easier to apply than the first derivative test Theorem 4 (Second Derivative Test) Let f be a function defined on an interval I
and c β I |
1 | 2886-2889 | We shall now give another test to examine local maxima and local minima of a
given function This test is often easier to apply than the first derivative test Theorem 4 (Second Derivative Test) Let f be a function defined on an interval I
and c β I Let f be twice differentiable at c |
1 | 2887-2890 | This test is often easier to apply than the first derivative test Theorem 4 (Second Derivative Test) Let f be a function defined on an interval I
and c β I Let f be twice differentiable at c Then
(i)
x = c is a point of local maxima if f β²(c) = 0 and f β³(c) < 0
The value f (c) is local maximum value of f |
1 | 2888-2891 | Theorem 4 (Second Derivative Test) Let f be a function defined on an interval I
and c β I Let f be twice differentiable at c Then
(i)
x = c is a point of local maxima if f β²(c) = 0 and f β³(c) < 0
The value f (c) is local maximum value of f (ii)
x = c is a point of local minima if
( )
0
f
β²c =
and f β³(c) > 0
In this case, f (c) is local minimum value of f |
1 | 2889-2892 | Let f be twice differentiable at c Then
(i)
x = c is a point of local maxima if f β²(c) = 0 and f β³(c) < 0
The value f (c) is local maximum value of f (ii)
x = c is a point of local minima if
( )
0
f
β²c =
and f β³(c) > 0
In this case, f (c) is local minimum value of f (iii)
The test fails if f β²(c) = 0 and f β³(c) = 0 |
1 | 2890-2893 | Then
(i)
x = c is a point of local maxima if f β²(c) = 0 and f β³(c) < 0
The value f (c) is local maximum value of f (ii)
x = c is a point of local minima if
( )
0
f
β²c =
and f β³(c) > 0
In this case, f (c) is local minimum value of f (iii)
The test fails if f β²(c) = 0 and f β³(c) = 0 In this case, we go back to the first derivative test and find whether c is a point of
local maxima, local minima or a point of inflexion |
1 | 2891-2894 | (ii)
x = c is a point of local minima if
( )
0
f
β²c =
and f β³(c) > 0
In this case, f (c) is local minimum value of f (iii)
The test fails if f β²(c) = 0 and f β³(c) = 0 In this case, we go back to the first derivative test and find whether c is a point of
local maxima, local minima or a point of inflexion ANote As f is twice differentiable at c, we mean
second order derivative of f exists at c |
1 | 2892-2895 | (iii)
The test fails if f β²(c) = 0 and f β³(c) = 0 In this case, we go back to the first derivative test and find whether c is a point of
local maxima, local minima or a point of inflexion ANote As f is twice differentiable at c, we mean
second order derivative of f exists at c Example 19 Find local minimum value of the function f
given by f (x) = 3 + |x|, x β R |
1 | 2893-2896 | In this case, we go back to the first derivative test and find whether c is a point of
local maxima, local minima or a point of inflexion ANote As f is twice differentiable at c, we mean
second order derivative of f exists at c Example 19 Find local minimum value of the function f
given by f (x) = 3 + |x|, x β R Solution Note that the given function is not differentiable
at x = 0 |
1 | 2894-2897 | ANote As f is twice differentiable at c, we mean
second order derivative of f exists at c Example 19 Find local minimum value of the function f
given by f (x) = 3 + |x|, x β R Solution Note that the given function is not differentiable
at x = 0 So, second derivative test fails |
1 | 2895-2898 | Example 19 Find local minimum value of the function f
given by f (x) = 3 + |x|, x β R Solution Note that the given function is not differentiable
at x = 0 So, second derivative test fails Let us try first
derivative test |
1 | 2896-2899 | Solution Note that the given function is not differentiable
at x = 0 So, second derivative test fails Let us try first
derivative test Note that 0 is a critical point of f |
1 | 2897-2900 | So, second derivative test fails Let us try first
derivative test Note that 0 is a critical point of f Now to
the left of 0, f (x) = 3 β x and so f β²(x) = β 1 < 0 |
1 | 2898-2901 | Let us try first
derivative test Note that 0 is a critical point of f Now to
the left of 0, f (x) = 3 β x and so f β²(x) = β 1 < 0 Also to
Rationalised 2023-24
APPLICATION OF DERIVATIVES
167
the right of 0, f (x) = 3 + x and so f β²(x) = 1 > 0 |
1 | 2899-2902 | Note that 0 is a critical point of f Now to
the left of 0, f (x) = 3 β x and so f β²(x) = β 1 < 0 Also to
Rationalised 2023-24
APPLICATION OF DERIVATIVES
167
the right of 0, f (x) = 3 + x and so f β²(x) = 1 > 0 Therefore, by first derivative test, x =
0 is a point of local minima of f and local minimum value of f is f (0) = 3 |
1 | 2900-2903 | Now to
the left of 0, f (x) = 3 β x and so f β²(x) = β 1 < 0 Also to
Rationalised 2023-24
APPLICATION OF DERIVATIVES
167
the right of 0, f (x) = 3 + x and so f β²(x) = 1 > 0 Therefore, by first derivative test, x =
0 is a point of local minima of f and local minimum value of f is f (0) = 3 Example 20 Find local maximum and local minimum values of the function f given by
f (x) = 3x4 + 4x3 β 12x2 + 12
Solution We have
f (x) = 3x4 + 4x3 β 12x2 + 12
or
f β²(x) = 12x3 + 12x2 β 24x = 12x (x β 1) (x + 2)
or
f β²(x) = 0 at x = 0, x = 1 and x = β 2 |
1 | 2901-2904 | Also to
Rationalised 2023-24
APPLICATION OF DERIVATIVES
167
the right of 0, f (x) = 3 + x and so f β²(x) = 1 > 0 Therefore, by first derivative test, x =
0 is a point of local minima of f and local minimum value of f is f (0) = 3 Example 20 Find local maximum and local minimum values of the function f given by
f (x) = 3x4 + 4x3 β 12x2 + 12
Solution We have
f (x) = 3x4 + 4x3 β 12x2 + 12
or
f β²(x) = 12x3 + 12x2 β 24x = 12x (x β 1) (x + 2)
or
f β²(x) = 0 at x = 0, x = 1 and x = β 2 Now
f β³(x) = 36x2 + 24x β 24 = 12 (3x2 + 2x β 2)
or
β²β²
= β
<
β²β²
=
>
β²β² β
=
>
ο£±


f
f
f
( )
( )
(
)
0
24
0
1
36
0
2
72
0
Therefore, by second derivative test, x = 0 is a point of local maxima and local
maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = β 2 are the points of local
minima and local minimum values of f at x = β 1 and β 2 are f (1) = 7 and f (β2) = β20,
respectively |
1 | 2902-2905 | Therefore, by first derivative test, x =
0 is a point of local minima of f and local minimum value of f is f (0) = 3 Example 20 Find local maximum and local minimum values of the function f given by
f (x) = 3x4 + 4x3 β 12x2 + 12
Solution We have
f (x) = 3x4 + 4x3 β 12x2 + 12
or
f β²(x) = 12x3 + 12x2 β 24x = 12x (x β 1) (x + 2)
or
f β²(x) = 0 at x = 0, x = 1 and x = β 2 Now
f β³(x) = 36x2 + 24x β 24 = 12 (3x2 + 2x β 2)
or
β²β²
= β
<
β²β²
=
>
β²β² β
=
>
ο£±


f
f
f
( )
( )
(
)
0
24
0
1
36
0
2
72
0
Therefore, by second derivative test, x = 0 is a point of local maxima and local
maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = β 2 are the points of local
minima and local minimum values of f at x = β 1 and β 2 are f (1) = 7 and f (β2) = β20,
respectively Example 21 Find all the points of local maxima and local minima of the function f
given by
f(x) = 2x3 β 6x2 + 6x +5 |
1 | 2903-2906 | Example 20 Find local maximum and local minimum values of the function f given by
f (x) = 3x4 + 4x3 β 12x2 + 12
Solution We have
f (x) = 3x4 + 4x3 β 12x2 + 12
or
f β²(x) = 12x3 + 12x2 β 24x = 12x (x β 1) (x + 2)
or
f β²(x) = 0 at x = 0, x = 1 and x = β 2 Now
f β³(x) = 36x2 + 24x β 24 = 12 (3x2 + 2x β 2)
or
β²β²
= β
<
β²β²
=
>
β²β² β
=
>
ο£±


f
f
f
( )
( )
(
)
0
24
0
1
36
0
2
72
0
Therefore, by second derivative test, x = 0 is a point of local maxima and local
maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = β 2 are the points of local
minima and local minimum values of f at x = β 1 and β 2 are f (1) = 7 and f (β2) = β20,
respectively Example 21 Find all the points of local maxima and local minima of the function f
given by
f(x) = 2x3 β 6x2 + 6x +5 Solution We have
f(x) = 2x3 β 6x2 + 6x +5
or
2
2
( )
6
12
6
6(
1)
( )
12(
1)
f
x
x
x
x
f
x
x
ο£± β²
=
β
+
=
β
 β²β²
=
β

Now f β²(x) = 0 gives x =1 |
1 | 2904-2907 | Now
f β³(x) = 36x2 + 24x β 24 = 12 (3x2 + 2x β 2)
or
β²β²
= β
<
β²β²
=
>
β²β² β
=
>
ο£±


f
f
f
( )
( )
(
)
0
24
0
1
36
0
2
72
0
Therefore, by second derivative test, x = 0 is a point of local maxima and local
maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = β 2 are the points of local
minima and local minimum values of f at x = β 1 and β 2 are f (1) = 7 and f (β2) = β20,
respectively Example 21 Find all the points of local maxima and local minima of the function f
given by
f(x) = 2x3 β 6x2 + 6x +5 Solution We have
f(x) = 2x3 β 6x2 + 6x +5
or
2
2
( )
6
12
6
6(
1)
( )
12(
1)
f
x
x
x
x
f
x
x
ο£± β²
=
β
+
=
β
 β²β²
=
β

Now f β²(x) = 0 gives x =1 Also f β³(1) = 0 |
1 | 2905-2908 | Example 21 Find all the points of local maxima and local minima of the function f
given by
f(x) = 2x3 β 6x2 + 6x +5 Solution We have
f(x) = 2x3 β 6x2 + 6x +5
or
2
2
( )
6
12
6
6(
1)
( )
12(
1)
f
x
x
x
x
f
x
x
ο£± β²
=
β
+
=
β
 β²β²
=
β

Now f β²(x) = 0 gives x =1 Also f β³(1) = 0 Therefore, the second derivative test
fails in this case |
1 | 2906-2909 | Solution We have
f(x) = 2x3 β 6x2 + 6x +5
or
2
2
( )
6
12
6
6(
1)
( )
12(
1)
f
x
x
x
x
f
x
x
ο£± β²
=
β
+
=
β
 β²β²
=
β

Now f β²(x) = 0 gives x =1 Also f β³(1) = 0 Therefore, the second derivative test
fails in this case So, we shall go back to the first derivative test |
1 | 2907-2910 | Also f β³(1) = 0 Therefore, the second derivative test
fails in this case So, we shall go back to the first derivative test We have already seen (Example 18) that, using first derivative test, x =1 is neither
a point of local maxima nor a point of local minima and so it is a point of inflexion |
1 | 2908-2911 | Therefore, the second derivative test
fails in this case So, we shall go back to the first derivative test We have already seen (Example 18) that, using first derivative test, x =1 is neither
a point of local maxima nor a point of local minima and so it is a point of inflexion Example 22 Find two positive numbers whose sum is 15 and the sum of whose
squares is minimum |
1 | 2909-2912 | So, we shall go back to the first derivative test We have already seen (Example 18) that, using first derivative test, x =1 is neither
a point of local maxima nor a point of local minima and so it is a point of inflexion Example 22 Find two positive numbers whose sum is 15 and the sum of whose
squares is minimum Solution Let one of the numbers be x |
1 | 2910-2913 | We have already seen (Example 18) that, using first derivative test, x =1 is neither
a point of local maxima nor a point of local minima and so it is a point of inflexion Example 22 Find two positive numbers whose sum is 15 and the sum of whose
squares is minimum Solution Let one of the numbers be x Then the other number is (15 β x) |
1 | 2911-2914 | Example 22 Find two positive numbers whose sum is 15 and the sum of whose
squares is minimum Solution Let one of the numbers be x Then the other number is (15 β x) Let S(x)
denote the sum of the squares of these numbers |
1 | 2912-2915 | Solution Let one of the numbers be x Then the other number is (15 β x) Let S(x)
denote the sum of the squares of these numbers Then
Rationalised 2023-24
MATHEMATICS
168
S(x) = x2 + (15 β x)2 = 2x2 β 30x + 225
or
S ( )
4
30
S ( )
4
x
x
x
β²
=
β
ο£±
ο£² β²β²
=
ο£³
Now Sβ²(x) = 0 gives
x =215 |
1 | 2913-2916 | Then the other number is (15 β x) Let S(x)
denote the sum of the squares of these numbers Then
Rationalised 2023-24
MATHEMATICS
168
S(x) = x2 + (15 β x)2 = 2x2 β 30x + 225
or
S ( )
4
30
S ( )
4
x
x
x
β²
=
β
ο£±
ο£² β²β²
=
ο£³
Now Sβ²(x) = 0 gives
x =215 Also
15
S
4
0
2
ο£Ά
β²β²
=
>

ο£·
ο£
ο£Έ |
1 | 2914-2917 | Let S(x)
denote the sum of the squares of these numbers Then
Rationalised 2023-24
MATHEMATICS
168
S(x) = x2 + (15 β x)2 = 2x2 β 30x + 225
or
S ( )
4
30
S ( )
4
x
x
x
β²
=
β
ο£±
ο£² β²β²
=
ο£³
Now Sβ²(x) = 0 gives
x =215 Also
15
S
4
0
2
ο£Ά
β²β²
=
>

ο£·
ο£
ο£Έ Therefore, by second derivative
test,
x =215
is the point of local minima of S |
1 | 2915-2918 | Then
Rationalised 2023-24
MATHEMATICS
168
S(x) = x2 + (15 β x)2 = 2x2 β 30x + 225
or
S ( )
4
30
S ( )
4
x
x
x
β²
=
β
ο£±
ο£² β²β²
=
ο£³
Now Sβ²(x) = 0 gives
x =215 Also
15
S
4
0
2
ο£Ά
β²β²
=
>

ο£·
ο£
ο£Έ Therefore, by second derivative
test,
x =215
is the point of local minima of S Hence the sum of squares of numbers is
minimum when the numbers are 15
2 and
15
15
15
2
2
β
= |
1 | 2916-2919 | Also
15
S
4
0
2
ο£Ά
β²β²
=
>

ο£·
ο£
ο£Έ Therefore, by second derivative
test,
x =215
is the point of local minima of S Hence the sum of squares of numbers is
minimum when the numbers are 15
2 and
15
15
15
2
2
β
= Remark Proceeding as in Example 34 one may prove that the two positive numbers,
whose sum is k and the sum of whose squares is minimum, are
2and
2
k
k |
1 | 2917-2920 | Therefore, by second derivative
test,
x =215
is the point of local minima of S Hence the sum of squares of numbers is
minimum when the numbers are 15
2 and
15
15
15
2
2
β
= Remark Proceeding as in Example 34 one may prove that the two positive numbers,
whose sum is k and the sum of whose squares is minimum, are
2and
2
k
k Example 23 Find the shortest distance of the point (0, c) from the parabola y = x2,
where 1
2 β€ c β€ 5 |
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