Find the local maxima and local minima, if any, of the following functions.

(i) $f(x)=x^{2}$

$\therefore f^{\prime}(x)=2 x$

Now,

$f^{\prime}(x)=0 \Rightarrow x=0$

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) $g(x)=x^{3}-3 x$

$\therefore g^{\prime}(x)=3 x^{2}-3$

Now,

$g^{\prime}(x)=0 \Rightarrow 3 x^{2}=3 \Rightarrow x=\pm 1$

$g^{\prime \prime}(x)=6 x$

$g^{\prime \prime}(1)=6>0$

$g^{\prime \prime}(-1)=-6<0$

By second derivative test, $x=1$ is a point of local minima and local minimum value of $g$ at $x=1$ is $g(1)=1^{3}-3=1-3=-2$. 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)=\sin x+\cos x, 0

$\therefore h^{\prime}(x)=\cos x-\sin x$

$h^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$

$h^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)$

$h^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0$

Therefore, by second derivative test, $x=\frac{\pi}{4}$ is a point of local maxima and the local maximum value of $h$ at $x=\frac{\pi}{4}$ is $h\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$.

(iv) $f(x)=\sin x-\cos x, 0

$\therefore f^{\prime}(x)=\cos x+\sin x$

$f^{\prime}(x)=0 \Rightarrow \cos x=-\sin x \Rightarrow \tan x=-1 \Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4} \in(0,2 \pi)$

$f^{\prime \prime}(x)=-\sin x+\cos x$

$f^{\prime \prime}\left(\frac{3 \pi}{4}\right)=-\sin \frac{3 \pi}{4}+\cos \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}<0$

$f^{\prime \prime}\left(\frac{7 \pi}{4}\right)=-\sin \frac{7 \pi}{4}+\cos \frac{7 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0$

Therefore, by second derivative test, $x=\frac{3 \pi}{4}$ is a point of local maxima and the local maximum value of $f$ at $x=\frac{3 \pi}{4}$ is

$f\left(\frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{4}-\cos \frac{3 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$

However, $x=\frac{7 \pi}{4}$ is a point of local minima and the local minimum value of $f$ at $x=\frac{7 \pi}{4}$ is $f\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{4}-\cos \frac{7 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}$.

(v) $f(x)=x^{3}-6 x^{2}+9 x+15$

$\therefore f^{\prime}(x)=3 x^{2}-12 x+9$

$f^{\prime}(x)=0 \Rightarrow 3\left(x^{2}-4 x+3\right)=0$

$\Rightarrow 3(x-1)(x-3)=0$

$\Rightarrow x=1,3$

Now, $f^{\prime \prime}(x)=6 x-12=6(x-2)$

$f^{\prime \prime}(1)=6(1-2)=-6<0$

$f^{\prime \prime}(3)=6(3-2)=6>0$

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) $g(x)=\frac{x}{2}+\frac{2}{x}, x>0$

$\therefore g^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}}$

Now,

$g^{\prime \prime}(x)=\frac{4}{x^{3}}$

$g^{\prime \prime}(2)=\frac{4}{2^{3}}=\frac{1}{2}>0$

Therefore, by second derivative test, $x=2$ is a point of local minima and the local minimum value of $g$ at $x=2$ is $g(2)=\frac{2}{2}+\frac{2}{2}=1+1=2$.

(vii) $g(x)=\frac{1}{x^{2}+2}$

$\therefore g^{\prime}(x)=\frac{-(2 x)}{\left(x^{2}+2\right)^{2}}$

$g^{\prime}(x)=0 \Rightarrow \frac{-2 x}{\left(x^{2}+2\right)^{2}}=0 \Rightarrow x=0$

Now, for values close to $x=0$ and to the left of $0, g^{\prime}(x)>0$. Also, for values close to $x=0$ and to the right of $0, g^{\prime}(x)<0$.

Therefore, by first derivative test, $x=0$ is a point of local maxima and the local maximum value of $g(0)$ is $\frac{1}{0+2}=\frac{1}{2} . .$

(viii) $f(x)=x \sqrt{1-x}, x>0$

$\therefore f^{\prime}(x)=\sqrt{1-x}+x \cdot \frac{1}{2 \sqrt{1-x}}(-1)=\sqrt{1-x}-\frac{x}{2 \sqrt{1-x}}$

$=\frac{2(1-x)-x}{2 \sqrt{1-x}}=\frac{2-3 x}{2 \sqrt{1-x}}$

$f^{\prime}(x)=0 \Rightarrow \frac{2-3 x}{2 \sqrt{1-x}}=0 \Rightarrow 2-3 x=0 \Rightarrow x=\frac{2}{3}$

$f^{\prime \prime}(x)=\frac{1}{2}\left[\frac{\sqrt{1-x}(-3)-(2-3 x)\left(\frac{-1}{2 \sqrt{1-x}}\right)}{1-x}\right]$

$=\frac{\sqrt{1-x}(-3)+(2-3 x)\left(\frac{1}{2 \sqrt{1-x}}\right)}{1}$

$=\frac{-6(1-x)+(2-3 x)}{4(1-x)^{\frac{3}{2}}}$

$=\frac{3 x-4}{4(1-x)^{\frac{3}{2}}}$

$f^{\prime \prime}\left(\frac{2}{3}\right)=\frac{3\left(\frac{2}{3}\right)-4}{4\left(1-\frac{2}{3}\right)^{\frac{3}{2}}}=\frac{2-4}{4\left(\frac{1}{3}\right)^{\frac{3}{2}}}=\frac{-1}{2\left(\frac{1}{3}\right)^{\frac{3}{2}}}<0$

Therefore, by second derivative test, $x=\frac{2}{3}$ is a point of local maxima and the local maximum value of $f$ at $x=\frac{2}{3}$ is

$f\left(\frac{2}{3}\right)=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9} .$