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Question:

(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 $[1,9]$

Solution:

(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.