Calculus solution Ex#4.4

Exercise 4.4

Question 7
Find the absolute maximum and minimum values
$f(x)=4 x^2-12 x+10 ;[1,2]$
Solution
Since it is a polynomial, therefore its continuous and differentiable everywhere.
To find critical points:

$f(x)=4 x^2-12 x+10-------(1)$

differentiating w.r.t $x$

$f^{\prime}(x)=8 x-12$

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

$\Rightarrow 8 x-12=0 \Rightarrow 8 x=12$

$\Rightarrow x=\frac{12}{8}=\frac{3}{2}$

The critical point $\frac{3}{2}$ lies in the interval $(1,2)$ .Therefore the absolute maximum and absolute minimum must occur at the end of points of the interval $[1,2]$ , or at the critical point $x=3/2$
Substituting $x=1,2 \quad\text{and } 3/2$ into (1).

$f(1) =4(1)^2-12(1)+10=4-12+10=2$

$f(2) =4(2)^2-12(2)+10=16-24+10=2$

$f(3 / 2) =4(3 / 2)^2-12(3 / 2)+10=4(9 / 4)=6(3)+10$

$=9-18+10=1$

Thus the absolute minimum of $f = 1$ at $x=3/2$ and absolute maximum of $f=2$ at $x=1,2$

Question 8

$f(x)=8 x-x^2 ;[0.6]$

Solution

$f(x)=8 x-x^2 --------------(1)$

The function $f(x)$ is continous and differentiable everywhere since its a polynomial
To find a critical point:

$f^{\prime}(x)=8-2 x$

set

$f^{\prime}(x)=0 \Rightarrow 8-2 x=0 \Rightarrow 8=2 x$

$\Rightarrow x=\frac{8 }{2}=4$

The critical point $4$ lies in the interval $[0,6]$ .Therefore the absolute maximum and absolute minimum must occur at the end of points of the interval $[0,6]$ , or at the critical point $x=4$
Substituting x=0,4,6 into (1)

$f(0)=8(0)-0=0$

$f(4)=8(4)-4^2=32-16=16$

$f(6)=8(6)-6^2=48-36=12$

Thus, the absolute minimum = $0$ at $x=0$ and absolute maximum= $16$ at $x=4$ .

Question 13
$f(x)=x-2 \sin x:[-\pi / 4, \pi / 2]$ .
Solution
The function is continuous and differentiable everywhere\\
To find the critical point:

$f(x)=x-2 \sin x -----------(1)$

differentiating w.r.t $x$

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

Set

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

$\Rightarrow 1-2 \cos x=0$

$\Rightarrow 2 \cos x=1$

$\Rightarrow \cos x=\frac{1}{2}$

$\Rightarrow x=\cos^{-1} (1 / 2)={\pi / 3}$

The critical point ${\pi / 3}$ lies in the interval $[-\pi / 4, \pi / 2]$ .Therefore the absolute maximum and absolute minimum must occur at the end of points of the interval $[-\pi / 4, \pi / 2]$ , or at the critical point $x={\pi / 3}$
substituting $x=-\pi/4, \quad \pi/3$ and $\pi/2$ into (1).

$f\left(\frac{-\pi}{4}\right) & =-\frac{\pi}{4}-2sin\left(\frac{-\pi}{4}\right)$

$&=-\frac{\pi}{4}+2sin\left(\frac{\pi}{4}\right)$

$& =\frac{-\pi}{4}+\frac{2}{\sqrt{2}}=\frac{-\pi}{4}+\sqrt{2}$

$f\left(\frac{\pi}{3}\right) & =\frac{\pi}{3}-2 \sin \left(\frac{\pi}{3}\right)$

$& =\frac{\pi}{3}-2\left(\frac{\sqrt{3}}{{2}}\right)=\frac{\pi}{3}-\sqrt{3$

$f\left(\frac{\pi}{2}\right) & =\frac{\pi}{2}-2 \sin \left(\frac{\pi}{2}\right)$

$& =\frac{\pi}{2}-2(1)=\frac{\pi}{2}-2$

Thus the maximum value of $f$ is $-\frac{\pi}{4}+\sqrt{2}$ at $x=-\pi / 4$ and the minimum value of $f$ is $\frac{\pi}{3}-\sqrt{3}$ at $x=\frac{\pi}{3}$ .

Question 16

$f(x)=|6-4x| ;[-3,3]$

Solution

$f(x)= \begin{cases} 6-4 x & 6-4 x \geq 0 \\ -(6-4x) & 6-4 x< 0\end{cases}\\$

$f(x)= \begin{cases} 6-4x & 6\geq 4x \\-6+4x) & 6< 4x\end{cases}\\$

$f(x)= \begin{cases} 6-4 x & x\leq \frac{3}{2} \\ -6+4x & x> \frac{3}{2}\end{cases} \quad (1)\\$

$f^{\prime}(x)= \begin{cases}-4 & x<3 / 2 \\ 4 & x>3 / 2\end{cases}\\$

We see that $f^\prime(x)$ does not exist at $x=\frac{3}{2}$ , therefore $x=\frac{3}{2}$ is the only critical point that lie on the interval $(-3,3)$ .The absolute maximum and absolute minimum value of $f$ must occur either at the end point of the interval $[-3,3]$ or at $x=\frac{3}{2}$ .
Substituting $-3$ , $\frac{3}{2}$ and $3$ into (1)