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 + cosx, 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
∴ g'(x) = 3x² − 3
Now,
g'(x) = 0 ⇒ 3x² = 3 ⇒ x = $\pm$1
g''(x) = 6x
g''(1) = 6 > 0
g''(−1) = −6 < 0
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 at is

(iv) f(x) = sin x − cos x, 0 < x < 2π
∴ $f\text{'}\left(x\right)=\cos x+\sin x$
$f\text{'}\left(x\right)=0\Rightarrow \cos x=-\sin x\Rightarrow \tan x=-1\Rightarrow x=\frac{3\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\in \left(0,2\mathrm{\pi }\right)$
$f\text{'}\text{'}\left(x\right)=-\sin x+\cos x$
$f\text{'}\text{'}\left(\frac{3\mathrm{\pi }}{4}\right)=-\sin \frac{3\mathrm{\pi }}{4}+\cos \frac{3\mathrm{\pi }}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}<0$
$f\text{'}\text{'}\left(\frac{7\mathrm{\pi }}{4}\right)=-\sin \frac{7\mathrm{\pi }}{4}+\cos \frac{7\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0$
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

$$\begin{gathered} \therefore f\text{'}\left(x\right)=3x^2-12x+9 \\ f\text{'}\left(x\right)=0\Rightarrow 3(x^2-4x+3)=0 \\ \Rightarrow 3(x-1)(x-3)=0 \\ \Rightarrow x=1,3 \\ \mathrm{Now},f\text{'}\text{'}\left(x\right)=6x-12=6(x-2) \\ f\text{'}\text{'}\left(1\right)=6(1-2)=-6<0 \\ f\text{'}\text{'}\left(3\right)=6(3-2)=6>0 \end{gathered}$$
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