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1
2818-2821
Note that the function f is increasing (i e , f β€²(x) > 0) in the interval (c – h, c) and decreasing (i e
1
2819-2822
e , f β€²(x) > 0) in the interval (c – h, c) and decreasing (i e , f β€²(x) < 0) in the interval (c, c + h)
1
2820-2823
, f β€²(x) > 0) in the interval (c – h, c) and decreasing (i e , f β€²(x) < 0) in the interval (c, c + h) This suggests that f β€²(c) must be zero
1
2821-2824
e , f β€²(x) < 0) in the interval (c, c + h) This suggests that f β€²(c) must be zero Fig 6
1
2822-2825
, f β€²(x) < 0) in the interval (c, c + h) This suggests that f β€²(c) must be zero Fig 6 11 Fig 6
1
2823-2826
This suggests that f β€²(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 β€²(x) < 0) in the interval (c – h, c) and increasing (i
1
2829-2832
Here f is decreasing (i e , f β€²(x) < 0) in the interval (c – h, c) and increasing (i e
1
2830-2833
e , f β€²(x) < 0) in the interval (c – h, c) and increasing (i e , f β€²(x) > 0) in the interval (c, c + h)
1
2831-2834
, f β€²(x) < 0) in the interval (c – h, c) and increasing (i e , f β€²(x) > 0) in the interval (c, c + h) This again suggest that f β€²(c) must be zero
1
2832-2835
e , f β€²(x) > 0) in the interval (c, c + h) This again suggest that f β€²(c) must be zero The above discussion lead us to the following theorem (without proof)
1
2833-2836
, f β€²(x) > 0) in the interval (c, c + h) This again suggest that f β€²(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 β€²(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 ∈ 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 ∈ I be any point If f has a local maxima or a local minima at x = c, then either f β€²(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 ∈ I be any point If f has a local maxima or a local minima at x = c, then either f β€²(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 ∈ I be any point If f has a local maxima or a local minima at x = c, then either f β€²(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 β€²(x) = 3x2 and so f β€²(0) = 0
1
2838-2841
If f has a local maxima or a local minima at x = c, then either f β€²(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 β€²(x) = 3x2 and so f β€²(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 β€²(x) = 3x2 and so f β€²(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 β€²(x) = 3x2 and so f β€²(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 β€²(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 β€²(c) = 0 or f is not differentiable is called a critical point of f Note that if f is continuous at c and f β€²(c) = 0, then there exists an h > 0 such that f is differentiable in the interval (c – h, c + h)
1
2842-2845
13) ANote A point c in the domain of a function f at which either f β€²(c) = 0 or f is not differentiable is called a critical point of f Note that if f is continuous at c and f β€²(c) = 0, then there exists an h > 0 such that f is differentiable in the interval (c – 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 β€²(c) = 0 or f is not differentiable is called a critical point of f Note that if f is continuous at c and f β€²(c) = 0, then there exists an h > 0 such that f is differentiable in the interval (c – 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 β€²(c) = 0, then there exists an h > 0 such that f is differentiable in the interval (c – 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 β€²(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 β€²(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 β€²(x) changes sign from positive to negative as x increases through c, i e , if f β€²(x) > 0 at every point sufficiently close to and to the left of c, and f β€²(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 β€²(x) changes sign from positive to negative as x increases through c, i e , if f β€²(x) > 0 at every point sufficiently close to and to the left of c, and f β€²(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 β€²(x) changes sign from negative to positive as x increases through c, i
1
2849-2852
e , if f β€²(x) > 0 at every point sufficiently close to and to the left of c, and f β€²(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 β€²(x) changes sign from negative to positive as x increases through c, i e
1
2850-2853
, if f β€²(x) > 0 at every point sufficiently close to and to the left of c, and f β€²(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 β€²(x) changes sign from negative to positive as x increases through c, i e , if f β€²(x) < 0 at every point sufficiently close to and to the left of c, and f β€²(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 β€²(x) changes sign from negative to positive as x increases through c, i e , if f β€²(x) < 0 at every point sufficiently close to and to the left of c, and f β€²(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 β€²(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 β€²(x) < 0 at every point sufficiently close to and to the left of c, and f β€²(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 β€²(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 β€²(x) < 0 at every point sufficiently close to and to the left of c, and f β€²(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 β€²(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 β€²(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 – 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 – 3x + 3 Solution We have f (x) = x3 – 3x + 3 or f β€²(x) = 3x2 – 3 = 3(x – 1) (x + 1) or f β€²(x) = 0 at x = 1 and x = – 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 – 3x + 3 Solution We have f (x) = x3 – 3x + 3 or f β€²(x) = 3x2 – 3 = 3(x – 1) (x + 1) or f β€²(x) = 0 at x = 1 and x = – 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 – 3x + 3 Solution We have f (x) = x3 – 3x + 3 or f β€²(x) = 3x2 – 3 = 3(x – 1) (x + 1) or f β€²(x) = 0 at x = 1 and x = – 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 β€²(x) > 0 and for values close to 1 and to the left of 1, f β€²(x) < 0
1
2864-2867
Solution We have f (x) = x3 – 3x + 3 or f β€²(x) = 3x2 – 3 = 3(x – 1) (x + 1) or f β€²(x) = 0 at x = 1 and x = – 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 β€²(x) > 0 and for values close to 1 and to the left of 1, f β€²(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 β€²(x) > 0 and for values close to 1 and to the left of 1, f β€²(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 = –1, note that f β€²(x) > 0, for values close to and to the left of –1 and f β€²(x) < 0, for values close to and to the right of – 1
1
2866-2869
Note that for values close to 1 and to the right of 1, f β€²(x) > 0 and for values close to 1 and to the left of 1, f β€²(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 = –1, note that f β€²(x) > 0, for values close to and to the left of –1 and f β€²(x) < 0, for values close to and to the right of – 1 Therefore, by first derivative test, x = – 1 is a point of local maxima and local maximum value is f (–1) = 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 = –1, note that f β€²(x) > 0, for values close to and to the left of –1 and f β€²(x) < 0, for values close to and to the right of – 1 Therefore, by first derivative test, x = – 1 is a point of local maxima and local maximum value is f (–1) = 5 Values of x Sign of f β€²β€²β€²β€²β€²(x) = 3(x – 1) (x + 1) Close to 1 to the right (say 1
1
2868-2871
In the case of x = –1, note that f β€²(x) > 0, for values close to and to the left of –1 and f β€²(x) < 0, for values close to and to the right of – 1 Therefore, by first derivative test, x = – 1 is a point of local maxima and local maximum value is f (–1) = 5 Values of x Sign of f β€²β€²β€²β€²β€²(x) = 3(x – 1) (x + 1) Close to 1 to the right (say 1 1 etc
1
2869-2872
Therefore, by first derivative test, x = – 1 is a point of local maxima and local maximum value is f (–1) = 5 Values of x Sign of f β€²β€²β€²β€²β€²(x) = 3(x – 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 β€²β€²β€²β€²β€²(x) = 3(x – 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 –1 to the right (say 0
1
2872-2875
) >0 to the left (say 0 9 etc ) <0 Close to –1 to the right (say 0 9 etc
1
2873-2876
9 etc ) <0 Close to –1 to the right (say 0 9 etc ) 0 to the left (say 1
1
2874-2877
) <0 Close to –1 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 βˆ’ < βˆ’ > Rationalised 2023-24 MATHEMATICS 166 Fig 6
1
2876-2879
) 0 to the left (say 1 1 etc ) 0 βˆ’ < βˆ’ > 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 – 6x2 + 6x +5
1
2877-2880
1 etc ) 0 βˆ’ < βˆ’ > 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 – 6x2 + 6x +5 Solution We have f (x) = 2x3 – 6x2 + 6x + 5 or f β€²(x) = 6x2 – 12x + 6 = 6(x – 1)2 or f β€²(x) = 0 at x = 1 Thus, x = 1 is the only critical point of f
1
2878-2881
) 0 βˆ’ < βˆ’ > 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 – 6x2 + 6x +5 Solution We have f (x) = 2x3 – 6x2 + 6x + 5 or f β€²(x) = 6x2 – 12x + 6 = 6(x – 1)2 or f β€²(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 – 6x2 + 6x +5 Solution We have f (x) = 2x3 – 6x2 + 6x + 5 or f β€²(x) = 6x2 – 12x + 6 = 6(x – 1)2 or f β€²(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 β€²(x) β‰₯ 0, for all x ∈ R and in particular f β€²(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 – 6x2 + 6x + 5 or f β€²(x) = 6x2 – 12x + 6 = 6(x – 1)2 or f β€²(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 β€²(x) β‰₯ 0, for all x ∈ R and in particular f β€²(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 β€²(x) β‰₯ 0, for all x ∈ R and in particular f β€²(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 β€²(x) β‰₯ 0, for all x ∈ R and in particular f β€²(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 β€²(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 β€²(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 β€²(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 β€²(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 ∈ 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 ∈ 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 ∈ I Let f be twice differentiable at c Then (i) x = c is a point of local maxima if f β€²(c) = 0 and f β€³(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 ∈ I Let f be twice differentiable at c Then (i) x = c is a point of local maxima if f β€²(c) = 0 and f β€³(c) < 0 The value f (c) is local maximum value of f (ii) x = c is a point of local minima if ( ) 0 f β€²c = and f β€³(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 β€²(c) = 0 and f β€³(c) < 0 The value f (c) is local maximum value of f (ii) x = c is a point of local minima if ( ) 0 f β€²c = and f β€³(c) > 0 In this case, f (c) is local minimum value of f (iii) The test fails if f β€²(c) = 0 and f β€³(c) = 0
1
2890-2893
Then (i) x = c is a point of local maxima if f β€²(c) = 0 and f β€³(c) < 0 The value f (c) is local maximum value of f (ii) x = c is a point of local minima if ( ) 0 f β€²c = and f β€³(c) > 0 In this case, f (c) is local minimum value of f (iii) The test fails if f β€²(c) = 0 and f β€³(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 β€²c = and f β€³(c) > 0 In this case, f (c) is local minimum value of f (iii) The test fails if f β€²(c) = 0 and f β€³(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 β€²(c) = 0 and f β€³(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 ∈ 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 ∈ 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 ∈ 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 ∈ 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 – x and so f β€²(x) = – 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 – x and so f β€²(x) = – 1 < 0 Also to Rationalised 2023-24 APPLICATION OF DERIVATIVES 167 the right of 0, f (x) = 3 + x and so f β€²(x) = 1 > 0
1
2899-2902
Note that 0 is a critical point of f Now to the left of 0, f (x) = 3 – x and so f β€²(x) = – 1 < 0 Also to Rationalised 2023-24 APPLICATION OF DERIVATIVES 167 the right of 0, f (x) = 3 + x and so f β€²(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 – x and so f β€²(x) = – 1 < 0 Also to Rationalised 2023-24 APPLICATION OF DERIVATIVES 167 the right of 0, f (x) = 3 + x and so f β€²(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 – 12x2 + 12 Solution We have f (x) = 3x4 + 4x3 – 12x2 + 12 or f β€²(x) = 12x3 + 12x2 – 24x = 12x (x – 1) (x + 2) or f β€²(x) = 0 at x = 0, x = 1 and x = – 2
1
2901-2904
Also to Rationalised 2023-24 APPLICATION OF DERIVATIVES 167 the right of 0, f (x) = 3 + x and so f β€²(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 – 12x2 + 12 Solution We have f (x) = 3x4 + 4x3 – 12x2 + 12 or f β€²(x) = 12x3 + 12x2 – 24x = 12x (x – 1) (x + 2) or f β€²(x) = 0 at x = 0, x = 1 and x = – 2 Now f β€³(x) = 36x2 + 24x – 24 = 12 (3x2 + 2x – 2) or β€²β€² = βˆ’ < β€²β€² = > β€²β€² βˆ’ = > 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 = – 2 are the points of local minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20, 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 – 12x2 + 12 Solution We have f (x) = 3x4 + 4x3 – 12x2 + 12 or f β€²(x) = 12x3 + 12x2 – 24x = 12x (x – 1) (x + 2) or f β€²(x) = 0 at x = 0, x = 1 and x = – 2 Now f β€³(x) = 36x2 + 24x – 24 = 12 (3x2 + 2x – 2) or β€²β€² = βˆ’ < β€²β€² = > β€²β€² βˆ’ = > 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 = – 2 are the points of local minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20, respectively Example 21 Find all the points of local maxima and local minima of the function f given by f(x) = 2x3 – 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 – 12x2 + 12 Solution We have f (x) = 3x4 + 4x3 – 12x2 + 12 or f β€²(x) = 12x3 + 12x2 – 24x = 12x (x – 1) (x + 2) or f β€²(x) = 0 at x = 0, x = 1 and x = – 2 Now f β€³(x) = 36x2 + 24x – 24 = 12 (3x2 + 2x – 2) or β€²β€² = βˆ’ < β€²β€² = > β€²β€² βˆ’ = > 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 = – 2 are the points of local minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20, respectively Example 21 Find all the points of local maxima and local minima of the function f given by f(x) = 2x3 – 6x2 + 6x +5 Solution We have f(x) = 2x3 – 6x2 + 6x +5 or 2 2 ( ) 6 12 6 6( 1) ( ) 12( 1) f x x x x f x x ο£± β€² = βˆ’ + = βˆ’  β€²β€² = βˆ’  Now f β€²(x) = 0 gives x =1
1
2904-2907
Now f β€³(x) = 36x2 + 24x – 24 = 12 (3x2 + 2x – 2) or β€²β€² = βˆ’ < β€²β€² = > β€²β€² βˆ’ = > 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 = – 2 are the points of local minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20, respectively Example 21 Find all the points of local maxima and local minima of the function f given by f(x) = 2x3 – 6x2 + 6x +5 Solution We have f(x) = 2x3 – 6x2 + 6x +5 or 2 2 ( ) 6 12 6 6( 1) ( ) 12( 1) f x x x x f x x ο£± β€² = βˆ’ + = βˆ’  β€²β€² = βˆ’  Now f β€²(x) = 0 gives x =1 Also f β€³(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 – 6x2 + 6x +5 Solution We have f(x) = 2x3 – 6x2 + 6x +5 or 2 2 ( ) 6 12 6 6( 1) ( ) 12( 1) f x x x x f x x ο£± β€² = βˆ’ + = βˆ’  β€²β€² = βˆ’  Now f β€²(x) = 0 gives x =1 Also f β€³(1) = 0 Therefore, the second derivative test fails in this case
1
2906-2909
Solution We have f(x) = 2x3 – 6x2 + 6x +5 or 2 2 ( ) 6 12 6 6( 1) ( ) 12( 1) f x x x x f x x ο£± β€² = βˆ’ + = βˆ’  β€²β€² = βˆ’  Now f β€²(x) = 0 gives x =1 Also f β€³(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 β€³(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 – 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 – 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 – x) Let S(x) denote the sum of the squares of these numbers Then Rationalised 2023-24 MATHEMATICS 168 S(x) = x2 + (15 – x)2 = 2x2 – 30x + 225 or S ( ) 4 30 S ( ) 4 x x x β€² = βˆ’ ο£± ο£² β€²β€² = ο£³ Now Sβ€²(x) = 0 gives x =215
1
2913-2916
Then the other number is (15 – x) Let S(x) denote the sum of the squares of these numbers Then Rationalised 2023-24 MATHEMATICS 168 S(x) = x2 + (15 – x)2 = 2x2 – 30x + 225 or S ( ) 4 30 S ( ) 4 x x x β€² = βˆ’ ο£± ο£² β€²β€² = ο£³ Now Sβ€²(x) = 0 gives x =215 Also 15 S 4 0 2 ο£Ά β€²β€² = >  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 – x)2 = 2x2 – 30x + 225 or S ( ) 4 30 S ( ) 4 x x x β€² = βˆ’ ο£± ο£² β€²β€² = ο£³ Now Sβ€²(x) = 0 gives x =215 Also 15 S 4 0 2 ο£Ά β€²β€² = >  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 – x)2 = 2x2 – 30x + 225 or S ( ) 4 30 S ( ) 4 x x x β€² = βˆ’ ο£± ο£² β€²β€² = ο£³ Now Sβ€²(x) = 0 gives x =215 Also 15 S 4 0 2 ο£Ά β€²β€² = >  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 βˆ’ =
1
2916-2919
Also 15 S 4 0 2 ο£Ά β€²β€² = >  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 βˆ’ = 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 βˆ’ = 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 ≀ c ≀ 5