Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

Solution:

(i) The given function is $f(x)=x^{3}$.

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

Now,

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

Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [−2, 2].

f(0) = 0

$f(-2)=(-2)^{3}=-8$

$f(2)=(2)^{3}=8$

Hence, we can conclude that the absolute maximum value of f on [−2, 2] is 8 occurring at x = 2. Also, the absolute minimum value of f on [−2, 2] is −8 occurring at x = −2.

(ii) The given function is f(x) = sin x + cos x.

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

Now,

$f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}$

Then, we evaluate the value of $f$ at critical point $x=\frac{\pi}{4}$ and at the end points of the interval $[0, \pi]$.

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

$f(0)=\sin 0+\cos 0=0+1=1$

$f(\pi)=\sin \pi+\cos \pi=0-1=-1$

Hence, we can conclude that the absolute maximum value of $f$ on $[0, \pi]$ is $\sqrt{2}$ occurring at $x=\frac{\pi}{4}$ and the absolute minimum value of $f$ on $[0, \pi]$ is $-1$ occurring at x = π.

(iii) The given function is $f(x)=4 x-\frac{1}{2} x^{2}$.

$\therefore f^{\prime}(x)=4-\frac{1}{2}(2 x)=4-x$

Now,

$f^{\prime}(x)=0 \Rightarrow x=4$

Then, we evaluate the value of $f$ at critical point $x=4$ and at the end points of the interval $\left[-2, \frac{9}{2}\right]$.

$f(4)=16-\frac{1}{2}(16)=16-8=8$

$f(-2)=-8-\frac{1}{2}(4)=-8-2=-10$

$f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^{2}=18-\frac{81}{8}=18-10.125=7.875$

Hence, we can conclude that the absolute maximum value of $f$ on $\left[-2, \frac{9}{2}\right]$ is 8 occurring at $x=4$ and the absolute minimum value of $f$ on $\left[-2, \frac{9}{2}\right]$ is $-10$ occurring at x = −2.

(iv) The given function is $f(x)=(x-1)^{2}+3$.

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

Now,

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

Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval [−3, 1].

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

$f(-3)=(-3-1)^{2}+3=16+3=19$

Hence, we can conclude that the absolute maximum value of f on [−3, 1] is 19 occurring at x = −3 and the minimum value of f on [−3, 1] is 3 occurring at x = 1.