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)
$$ f(-3)=+6-4(-3)=+6+12=+18$$
$$f\left(\frac{3}{2}\right)=6-4\left(\frac{3}{2}\right)=6-6=0$$
$$f(3)=6+4(3)=-6+12=+6.$$
Thus absolute minimum value of $f$ is 0 at $x=3/2$ and absolute maximum value is 18 at $x=-3$.