If the solve the problem

(i)

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

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

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

$=x^{3}-5 x^{2}+8 x-4$

$\Rightarrow f^{\prime}(x)=3 x^{2}-10 x+8$

For the local maxima or minima, we must have

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

$\Rightarrow 3 x^{2}-10 x+8=0$

$\Rightarrow 3 x^{2}-6 x-4 x+8=0$

$\Rightarrow(x-2)(3 x-4)=0$

$\Rightarrow x=2$ and $\frac{4}{3}$

Thus, $x=2$ and $x=\frac{4}{3}$ are the possible points of local maxima or local minima.

Now,

$f^{\prime \prime}(x)=6 x-10$

At $x=2:$

$f^{\prime \prime}(2)=6(2)-10=2>0$

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

The local minimum value is given by

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

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

The local minimum value is given by

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

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

The local minimum value is given by

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

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

$f^{\prime \prime}\left(\frac{4}{3}\right)=6\left(\frac{4}{3}\right)-10=-2<0$

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

The local maximum value is given by

$f\left(\frac{4}{3}\right)=\left(\frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)^{2}=\frac{1}{3} \times \frac{4}{9}=\frac{4}{27}$

(ii)

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

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

For the local maxima or minima, we must have

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

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

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

$\Rightarrow 2-2 x=x$

$\Rightarrow 3 x=2$

$\Rightarrow x=\frac{2}{3}$

Thus, $x=\frac{2}{3}$ is the possible point of local maxima or local minima.

Now,

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

At $x=\frac{2}{3}:$

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

So, $x=\frac{2}{3}$ is the point of local maximum.

The local maximum value is given by

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

(iii)

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

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

For the local maxima or minima, we must have

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

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

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

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

$\Rightarrow(x-1)^{2}(x+1)[-5 x-1]=0$

$\Rightarrow x=1,-1$ and $\frac{-1}{5}$

Thus, $x=1, x=-1$ and $x=\frac{-1}{5}$ are the possible points of local maxima or local minima.

Now,

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

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

At $x=1:$

$f^{\prime \prime}(1)=-6(1-1)(1+1)^{2}+6(1+1)(1-1)^{2}-2(1-1)^{3}-6(1-1)^{2}(1+1)=0$

So, it is a point of inflexion.

At $x=-1:$

$f^{\prime \prime}(-1)=-6(-1-1)(-1+1)^{2}+6(-1+1)(-1-1)^{2}-2(-1-1)^{3}-6(-1-1)^{2}(-1+1)=16>0$

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

The local minimum value is given by

$f(-1)=-(1-1)^{3}(-1+1)^{2}=0$

At $x=-\frac{1}{5}:$

$f^{\prime \prime}\left(-\frac{1}{5}\right)=-6\left(-\frac{1}{5}-1\right)\left(-\frac{1}{5}+1\right)^{2}+6\left(-\frac{1}{5}+1\right)\left(-\frac{1}{5}-1\right)^{2}+2\left(-\frac{1}{5}-1\right)^{3}-6\left(-\frac{1}{5}-1\right)^{2}\left(-\frac{1}{5}+1\right)$

$=\frac{576}{125}+\frac{384}{125}-\frac{432}{125}-\frac{864}{125}=\frac{-336}{125}<0$

So, $x=-\frac{1}{5}$ is the point of local maximum.

The local maximum value is given by

$f\left(-\frac{1}{5}\right)=-\left(-\frac{1}{5}-1\right)^{3}\left(-\frac{1}{5}+1\right)^{2}=-\left(\frac{-216}{125}\right)\left(\frac{16}{25}\right)=\frac{3465}{3125}$