Question

Find the global maximum and minimum values of the following function on the given interval. If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use.

f(x)=4e−x−4e−2x  [0,1]

Global maxima

x = ____ Y=___

x=_____ Y=____

x=____ Y=___

Global minima

x=____ Y=___

x=____ Y=___

x=___ Y=___

x=____ Y=___

Answer 1

$$f(x)=4e^{−x}−4e^{−2x} \qquad[0,1]\\ \begin{aligned} f'(x) =& -4e^{-x}+8e^{-2x} \end{aligned}$$

Find the critical points:

$$\begin{aligned} f'(x)=0 \implies -4e^{-x}+8e^{-2x}&=0\\ 8e^{-2x}&=4e^{-x}\\ 2e^{-2x}&=e^{-x}\\ 2&=e^{-x}/e^{-2x}\\ 2&=e^{x}\\ \ln(2) &=x \end{aligned}$$

To find the absolute maximum and absolute minimum, then, we evaluate $f$ at the critical point and on the endpoints of the interval:

$$\begin{aligned} f(\ln(2))=&4e^{−\ln(2)}−4e^{−2\ln(2)}\\ =&4(\tfrac{1}{2})-4(\tfrac{1}{4})\\ =&2-1\\ =&1 \end{aligned}$$

$$\begin{aligned} f(0)=&4e^{0}−4e^{0}\\ =&4-4\\ =&0 \end{aligned}$$

$$\begin{aligned} f(1)=&4e^{−1}−4e^{−2}\\ =&4(0.3679)-4(0.1353)\\ =&1.4716-0.5412\\ \approx&0.93 \end{aligned}$$

Therefore, $f$ achieves its absolute minimum of at $x=0$ and its absolute maximum of $1$ at $x= \ln(2)$

Thus:

Absolute maxima = $1$

Absolute minima =