(i) $f(x)=4 x-\frac{x^{2}}{2}$ in $[-2,4,5]$
(ii) $f(x)=(x-1)^{2}+3$ in $[-3,1]$
(iii) $f(x)=3 x 4-8 x^{3}+12 x^{2}-48 x+25$ in $[0,3]$
(iv) $f(x)=(x-2) \sqrt{x-1}$ in $[1,9]$
(i)
Given: $f(x)=4 x-\frac{x^{2}}{2}$
$\Rightarrow f^{\prime}(x)=4-x$
For a local maximum or a local minimum, we must have
$f^{\prime}(x)=0$
$\Rightarrow 4-x=0$
$\Rightarrow x=4$
Thus, the critical points of $f$ are $-2,4$ and $4.5$.
Now,
$f(-2)=4(-2)-\frac{(-2)^{2}}{2}=-8-2=-10$
$f(4)=4(4)-\frac{(4)^{2}}{2}=16-8=8$
$f(4.5)=4(4.5)-\frac{(4.5)^{2}}{2}=18-10.125=7.875$
Hence, the absolute maximum value when $x=4$ is 8 and the absolute minimum value when $x=-2$ is $-10$.
(ii)
Given : $f(x)=(x-1)^{2}+3$
$\Rightarrow f^{\prime}(x)=2(x-1)$
For a local maximum or a local minimum, we must have
$f^{\prime}(x)=0$
$\Rightarrow 2(x-1)=0$
$\Rightarrow x=1$
Thus, the critical points of $f$ are $-3$ and 1 .
Now,
$f(-3)=(-3-1)^{2}+3=16+3=19$
$f(1)=(1-1)^{2}+3=3$
Hence, the absolute maximum value when $x=-3$ is 19 and the absolute minimum value when $x=1$ is $3 .$
(iii)
Given : $f(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+25$
$\Rightarrow f^{\prime}(x)=12 x^{3}-24 x^{2}+24 x-48$
For a local maximum or a local minimum, we must have
$f^{\prime}(x)=0$
$\Rightarrow 12 x^{3}-24 x^{2}+24 x-48=0$
$\Rightarrow x^{3}-2 x^{2}+2 x-4=0$
$\Rightarrow x^{2}(x-2)+2(x-2)=0$
$\Rightarrow(x-2)\left(x^{2}+2\right)=0$
$\Rightarrow x-2=0$ or $\left(x^{2}+2\right)=0$
$\Rightarrow x=2$
No real root exists for $\left(x^{2}+2\right)=0$.
Thus, the critical points of $f$ are 0,2 and 3 .
Now,
$f(0)=3(0)^{4}-8(0)^{3}+12(0)^{2}-48(0)+25=25$
$f(2)=3(2)^{4}-8(2)^{3}+12(2)^{2}-48(2)+25=-39$
$f(3)=3(3)^{4}-8(3)^{3}+12(3)^{2}-48(3)+25=16$
Hence, the absolute maximum value when $x=0$ is 25 and the absolute minimum value when $x=2$ is $-39 .$
(iv)
Given: $f(x)=(x-2) \sqrt{x-1}$
$\Rightarrow f^{\prime}(x)=\sqrt{x-1}+\frac{(x-2)}{2 \sqrt{x-1}}$
For a local maximum or a local minimum, we must have
$f^{\prime}(x)=0$
$\Rightarrow \sqrt{x-1}+\frac{(x-2)}{2 \sqrt{x-1}}=0$
$\Rightarrow 2(x-1)+(x-2)=0$
$\Rightarrow 2 x-2+x-2=0$
$\Rightarrow 3 x-4=0$
$\Rightarrow 3 x=4$
$\Rightarrow x=\frac{4}{3}$
Thus, the critical points of $f$ are $1, \frac{4}{3}$ and $9 .$
Now,
$f(1)=(1-2) \sqrt{1-1}=0$
$f\left(\frac{4}{3}\right)=\left(\frac{4}{3}-2\right) \sqrt{\frac{4}{3}-1}=\frac{-2}{3} \times \frac{1}{\sqrt{3}}=-\frac{2}{3 \sqrt{3}}$
$f(9)=(9-2) \sqrt{9-1}=14 \sqrt{2}$
Hence, the absolute maximum value when $x=9$ is $14 \sqrt{2}$ and the absolute minimum value when $x=\frac{4}{3}$ is $-\frac{2}{3 \sqrt{3}} .$
Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.
(iv)
Given: $f(x)=(x-2) \sqrt{x-1}$
$\Rightarrow f^{\prime}(x)=\sqrt{x-1}+\frac{(x-2)}{2 \sqrt{x-1}}$
For a local maximum or a local minimum, we must have
$f^{\prime}(x)=0$
$\Rightarrow \sqrt{x-1}+\frac{(x-2)}{2 \sqrt{x-1}}=0$
$\Rightarrow 2(x-1)+(x-2)=0$
$\Rightarrow 2 x-2+x-2=0$
$\Rightarrow 3 x-4=0$
$\Rightarrow 3 x=4$
$\Rightarrow x=\frac{4}{3}$
Thus, the critical points of $f$ are $1, \frac{4}{3}$ and $9 .$
Now,
$f(1)=(1-2) \sqrt{1-1}=0$
$f\left(\frac{4}{3}\right)=\left(\frac{4}{3}-2\right) \sqrt{\frac{4}{3}-1}=\frac{-2}{3} \times \frac{1}{\sqrt{3}}=-\frac{2}{3 \sqrt{3}}$
$f(9)=(9-2) \sqrt{9-1}=14 \sqrt{2}$
Hence, the absolute maximum value when $x=9$ is $14 \sqrt{2}$ and the absolute minimum value when $x=\frac{4}{3}$ is $-\frac{2}{3 \sqrt{3}} .$
Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.