Question

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i). f(x) = x² (ii). g(x) = x³ − 3x

(iii). h(x) = sinx + cos, 0 < (iv). f(x) = sinx − cos x, 0 < x < 2π

(v). f(x) = x³ − 6x² + 9x + 15

(vi).

(vii).

(viii).

Answer 1

(i) f(x) = x²

Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.

We have, which is positive.

Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = 0.

(ii) g(x) = x³ − 3x

By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1 is g(1) = 1³ − 3 = 1 − 3 = −2. However,

x = −1 is a point of local maxima and local maximum value of g at

x = −1 is g(1) = (−1)³ − 3 (− 1) = − 1 + 3 = 2.

(iii) h(x) = sinx + cosx, 0 < x <

Therefore, by second derivative test,is a point of local maxima and the local maximum value of h atis

(iv) f(x) = sin x − cos x, 0 < x < 2π

Therefore, by second derivative test,is a point of local maxima and the local maximum value of f at is

However,is a point of local minima and the local minimum value of f at is .

(v) f(x) = x³ − 6x² + 9x + 15

Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is f(1) = 1 − 6 + 9 + 15 = 19. However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is f(3) = 27 − 54 + 27 + 15 = 15.

(vi)

Since x > 0, we take x = 2.

Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is g(2) =

(vii)

Now, for values close to x = 0 and to the left of 0,Also, for values close to x = 0 and to the right of 0,.

Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of.

(viii)

Therefore, by second derivative test,is a point of local maxima and the local maximum value of f at is