If the solve the problem

Solution:

(i)

Given : $f(x)=x^{4}-62 x^{2}+120 x+9$

$\Rightarrow f^{\prime}(x)=4 x^{3}-124 x+120$

For the local maxima or minima, we must have

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

$\Rightarrow 4 x^{3}-124 x+120=0$

$\Rightarrow x^{3}-31 x+30=0$

$\Rightarrow(x-1)\left(x^{2}+x-30\right)=0$

$\Rightarrow(x-1)(x+6)(x-5)=0$

$\Rightarrow x=1,5$ and $-6$

Thus, $x=1, x=5$ and $x=-6$ are the possible points of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=12 x^{2}-124$

At $x=1:$

$f^{\prime \prime}(1)=12(1)^{2}-124=-112<0$

So, $x=1$ is the point of local maximum.

The local maximum value is given by

$f(1)=1^{4}-62(1)^{2}+120 \times 1+9=68$

At $x=5:$

$f^{\prime \prime}(5)=12(5)^{2}-124=176>0$

So, $x=5$ is the point of local minimum.

The local minimum value is given by

$f(5)=5^{4}-62(5)^{2}+120 \times 5+9=-316$

At $x=-6:$

$f^{\prime \prime}(-6)=12(-6)^{2}-124=308>0$

So, $x=-6$ is the point of local maximum.

The local minimum value is given by

$f(-6)=(-6)^{4}-62(-6)^{2}+120 \times(-6)+9=-1647$

(ii)

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

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

For the local maxima or minima, we must have

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

$\Rightarrow 3 x^{2}-12 x+9=0$

$\Rightarrow x^{2}-4 x+3=0$

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

 

$\Rightarrow x=1$ and 3

Thus, $x=1$ and $x=3$ are the possible points of local maxima or local minima.

Now,

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

At $x=1:$

$f^{\prime \prime}(1)=6(1)-12=-6<0$

So, $x=1$ is the point of local maximum.

The local maximum value is given by

$f(1)=1^{3}-6(1)^{2}+9 \times 1+15=19$

At $x=3$ :

$f^{\prime \prime}(3)=6(3)-12=6>0$

So, $x=3$ is the point of local minimum.

The local minimum value is given by

$f(3)=3^{3}-6(3)^{2}+9 \times 3+15=15$

(iii)

Given : $f(x)=(x-1)(x+2)^{2}$

$=(x-1)\left(x^{2}+4 x+4\right)$

$=x^{3}+4 x^{2}+4 x-x^{2}-4 x-4$

$=x^{3}+3 x^{2}-4$

 

$\Rightarrow f^{\prime}(x)=3 x^{2}+6 x$

For the local maxima or minima, we must have

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

$\Rightarrow 3 x^{2}+6 x=0$

$\Rightarrow 3 x(x+2)=0$

 

$\Rightarrow x=0$ and $-2$

Thus, $x=0$ and $x=-2$ are the possible points of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=6 x+6$

At $x=0:$

$f^{\prime \prime}(0)=6(0)+6=6>0$

So, $x=0$ is the point of local minimum.

The local minimum value is given by

$f(0)=(0-1)(0+2)^{2}=-4$

At $x=-2:$

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

So, $x=-2$ is the point of local maximum.

The local maximum value is given by

$f(-2)=(-2-1)(-2+2)^{2}=0$

(iv)

Given : $f(x)=\frac{2}{x}-\frac{2}{x^{2}}=2 x^{-1}-2 x^{-2}$

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

For the local maxima or minima, we must have

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

$\Rightarrow \frac{4}{x^{3}}-\frac{2}{x^{2}}=0$

$\Rightarrow 4-2 x=0$

 

$\Rightarrow x=2$

Thus, $x=2$ is the possible point of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=\frac{-12}{x^{4}}+\frac{4}{x^{3}}$

At $x=2:$

$f^{\prime \prime}(2)=\frac{-12}{16}+\frac{4}{8}=\frac{-12+8}{16}=\frac{-1}{4}<0$

So, $x=2$ is the point of local maximum.

The local maximum value is given by

$f(2)=\frac{2}{2}-\frac{2}{2^{2}}=1-\frac{1}{2}=\frac{1}{2}$

(v)

Given: $f(x)=x e^{x}$

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

For the local maxima or minima, we must have

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

$\Rightarrow e^{x}+x e^{x}=0$

$\Rightarrow e^{x}(1+x)=0$

$\Rightarrow e^{x} \neq 0, x=-1$

 

$\Rightarrow x=-1$

Thus, $x=-1$ is the possible point of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=e^{x}+e^{x}+x e^{x}$

At $x=-1:$

$f^{\prime \prime}(-1)=e^{-1}+e^{-1}-e^{-1}=e^{-1}>0$

So, $x=-1$ is the point of local minimum.

The local minimum value is given by

$f(-1)=-e^{-1}=-\frac{1}{e}$

(vi)

Given : $f(x)=\frac{x}{2}+\frac{2}{x}$

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

For the local maxima or minima, we must have

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

$\Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0$

$\Rightarrow x^{2}=4$

 

$\Rightarrow x=2$ and $-2$

Thus, $x=2$ and $x=-2$ are the possible points of local maxima or a local minima.

Since $x>0, x=2$

Now,

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

At $x=2:$

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

So, $x=2$ is the point of local minimum.

The local minimum value is given by

$f(2)=\frac{x}{2}+\frac{2}{x}=1+1=2$

(vii)

Given : $f(x)=(x+1)(x+2)^{\frac{1}{3}}$

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

For the local maxima or minima, we must have

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

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

$\Rightarrow \frac{1}{3}(x+1)=-(x+2)^{\frac{1}{3}} \times(x+2)^{\frac{2}{3}}$

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

$\Rightarrow x+1=-3 x-6$

 

$\Rightarrow x=\frac{-7}{4}$

Thus, $x=\frac{-7}{4}$ is the possible point of local maxima or local minima.

Now,

$f^{\prime \prime}\left(\frac{-7}{4}\right)=\frac{2}{3}(x+2)^{\frac{-2}{3}}-\frac{2}{9}(x+1)(x+2)^{\frac{-5}{3}}$

At $x=\frac{-7}{4}:$

$f^{\prime \prime}\left(\frac{-7}{4}\right)=\frac{2}{3}\left(\frac{-7}{4}+2\right)^{\frac{-2}{3}}-\frac{2}{9}\left(\frac{-7}{4}+1\right)\left(\frac{-7}{4}+2\right)^{\frac{-5}{3}}=\frac{2}{3}\left(\frac{1}{4}\right)^{\frac{-2}{3}}+\frac{1}{18}\left(\frac{1}{4}\right)^{\frac{-5}{2}}>0$

So, $x=\frac{-7}{4}$ is the point of local minimum.

The local minimum value is given by

$f\left(\frac{-7}{4}\right)=\left(\frac{-7}{4}+1\right)\left(\frac{-7}{4}+2\right)^{\frac{1}{3}}=\frac{-3}{4}\left(\frac{1}{4}\right)^{\frac{1}{3}}=\frac{-3}{4^{\frac{4}{3}}}$

(viii)

Given : $f(x)=x \sqrt{32-x^{2}}$

$\Rightarrow f^{\prime}(x)=\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}}$

For the local maxima or minima, we must have

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

$\Rightarrow \sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}}=0$

 

$\Rightarrow \sqrt{32-x^{2}}=\frac{x}{\sqrt{32-x^{2}}}$

$\Rightarrow 32-x^{2}=x^{2}$

$\Rightarrow x^{2}=16$

 

$\Rightarrow x=\pm 4$

Thus, $x=4$ and $x=-4$ are the possible points of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=\frac{-x}{\sqrt{32-x^{2}}}-\left(\frac{2 x \sqrt{32-x^{2}}+\frac{x^{3}}{\sqrt{32-x^{2}}}}{32-x^{2}}\right)=\frac{-x}{\sqrt{32-x^{2}}}-\left(\frac{2 x\left(32-x^{2}\right)+x^{3}}{\left(32-x^{2}\right) \sqrt{32-x^{2}}}\right)$

At $x=4:$

$f^{\prime \prime}(4)=\frac{-4}{\sqrt{32-4^{2}}}-\left[\frac{8\left(32-4^{2}\right)+4^{3}}{\left(32-4^{2}\right) \sqrt{32-4^{2}}}\right]=-1-\frac{192}{64}=-3<0$

So, $x=4$ is the point of local maximum.

The local maximum value is given by

$f(4)=4 \sqrt{32-4^{2}}=16$

At $x=-4:$

$f^{\prime \prime}(-4)=\frac{4}{\sqrt{32-4^{2}}}+\left[\frac{8\left(32-4^{2}\right)-4^{3}}{\left(32-4^{2}\right) \sqrt{32-4^{2}}}\right]=1+2=3>0$

So, $x=-4$ is the point of local minimum.

The local minimum value is given by

$f(-4)=-4 \sqrt{32-4^{2}}=-16$

(ix) $f(x)=x^{3}-2 a x^{2}+a^{2} x$

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

For the local maxima or minima, we must have

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

$\Rightarrow 3 x^{2}-4 a x+a^{2}=0$

$\Rightarrow 3 x^{2}-3 a x-a x+a^{2}=0$

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

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

 

$\Rightarrow x=a$ and $\frac{a}{3}$

Thus, $x=a$ and $x=\frac{a}{3}$ are the possible points of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=6 x-4 a$

At $x=a:$

$f^{\prime \prime}(a)=6(a)-4 a=2 a>0$

So, $x=a$ is the point of local minimum.

The local minimum value is given by

$f(a)=a^{3}-2 a(a)^{2}+a^{2}(a)=0$

At $x=\frac{a}{3}:$

$f^{\prime \prime}\left(\frac{a}{3}\right)=6\left(\frac{a}{3}\right)-4 a=-2 a<0$

So, $x=\frac{a}{3}$ is the point of local maximum.

The local maximum value is given by

$f\left(\frac{a}{3}\right)=\left(\frac{a}{3}\right)^{3}-2 a\left(\frac{a}{3}\right)^{2}+a^{2}\left(\frac{a}{3}\right)=\frac{a^{3}}{27}-\frac{2 a^{3}}{9}+\frac{a^{3}}{3}=\frac{4 a^{3}}{27}$

(x)

Given : $f(x)=x+\frac{a^{2}}{x}$

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

For the local maxima or minima, we must have

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

$\Rightarrow 1-\frac{a^{2}}{x^{2}}=0$

$\Rightarrow x^{2}=a^{2}$

 

$\Rightarrow x=\pm a$

Thus, $x=a$ and $x=-a$ are the possible points of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=\frac{a^{2}}{x^{3}}$

At $x=a:$

$f^{\prime \prime}(a)=\frac{a^{2}}{(a)^{3}}=\frac{1}{a}>0$

So, $x=a$ is the point of local minimum.

The local minimum value is given by

$f(a)=x+\frac{a^{2}}{x}=a+a=2 a$

At $x=-a:$

$f^{\prime \prime}(a)=\frac{a^{2}}{(a)^{3}}=\frac{1}{a}>0$

So, $x=a$ is the point of local minimum.

The local minimum value isgiven by

$f(a)=x+\frac{a^{2}}{x}=a+a=2 a$

At $x=-a:$

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

So, $x=-a$ is the point of local maximum.

The local maximum value is given by

$f(-a)=x+\frac{a^{2}}{x}=-a-a=-2 a$

(xi)

Given: $f(x)=x \sqrt{2-x^{2}}$

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

For the local maxima or minima, we must have

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

$\Rightarrow \sqrt{2-x^{2}}-\frac{x^{2}}{\sqrt{2-x^{2}}}=0$

$\Rightarrow \sqrt{2-x^{2}}=\frac{x}{\sqrt{2-x^{2}}}$

$\Rightarrow 2-x^{2}=x^{2}$

$\Rightarrow x^{2}=1$

$\Rightarrow x=\pm 1$

Thus, $x=1$ and $x=-1$ are the possible points of local maxima or local minima.

Now,

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

At $x=1:$

$f^{\prime \prime}(1)=\frac{-1}{\sqrt{2-1^{2}}}-\left[\frac{2\left(2-1^{2}\right)+1^{3}}{\left(2-1^{2}\right) \sqrt{2-1^{2}}}\right]=-\frac{1}{2}-\frac{3}{2}=-2<0$

So, $x=1$ is the point of local maximum.

The local maximum value is given by

$f(4)=1 \sqrt{2-1^{2}}=1$

At $x=-1:$

$f^{\prime \prime}(-1)=\frac{1}{\sqrt{2-1^{2}}}+\left[\frac{2\left(2-1^{2}\right)-1^{3}}{\left(2-1^{2}\right) \sqrt{2-1^{2}}}\right]=1+1=2>0$

So, $x=-1$ is the point of local minimum.

The local minimum value is given by

$f(-1)=-1 \sqrt{2-1^{2}}=-1$

(xii) 

Given: $f(x)=x+\sqrt{1-x}$

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

For the local maxima or minima, we must have

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

$\Rightarrow 1-\frac{1}{2 \sqrt{1-x}}=0$

$\Rightarrow \sqrt{1-x}=\frac{1}{2}$

$\Rightarrow 1-x=\frac{1}{4}$

 

$\Rightarrow x=\frac{3}{4}$

Thus, $x=\frac{3}{4}$ is the possible point of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=-\frac{\frac{1}{4 \sqrt{1-x}}}{4(1-x)}$

At $x=\frac{3}{4}:$

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

So, $x=\frac{3}{4}$ is the point of local maximum.

The local maximum value is given by

$f\left(\frac{3}{4}\right)=\frac{3}{4}+\sqrt{1-\frac{3}{4}}=\frac{5}{4}$