Is #f(x)=(3-e^(2x))/x# increasing or decreasing at #x=-1#?

Question

Charles Arocha · Accepted Answer

#f# decreases at #x=-1#.

We know that, #f'(a) <0 rArr f" is "darr" at "x=a#. Now, #f(x)=(3-e^(2x))/x=(3-e^(2x))x^-1# To find #f'(x)#, we use the Product and Chain Rule. #f'(x)={3-e^(2x)}d/dx(x^-1)+x^-1{d/dx(3-e^(2x)}# #={3-e^(2x)}(-1x^-2)+x^-1{0-e^(2x)d/dx(2x)}# #=(e^(2x)-3)/x^2-(2e^(2x))/x# #:. f'(x)=(e^(2x)-2xe^(2x)-3)/x^2={(1-2x)e^(2x)-3}/x^2# #rArr f'(-1)={(1+2)e^-2-3}/(-1)^2=3(e^-2-1)# Since, #2ltelt3, 4lte^2lt9, 1/4gt1/e^2gt1/9,# we have, #f'(-1) <0.# Therefore, #f# decreases at #x=-1#. Have fun with math!

Ezekiel Alexander · Answer

undefined To determine if $ f(x) = \frac{{3 - e^{2x}}}{x} $ is increasing or decreasing at $ x = -1 $, we can use the first derivative test.

First, find the derivative of $ f(x) $:

$$ f'(x) = \frac{{d}}{{dx}} \left( \frac{{3 - e^{2x}}}{x} \right) $$

Using the quotient rule,

$$ f'(x) = \frac{{(x)(0) - (3 - e^{2x})(1)}}{{x^2}} = \frac{{e^{2x} - 3}}{{x^2}} $$

Now evaluate $ f'(-1) $:

$$ f'(-1) = \frac{{e^{-2} - 3}}{{(-1)^2}} = \frac{{e^{-2} - 3}}{1} = e^{-2} - 3 $$

Since $ e^{-2} $ is less than 1, $ e^{-2} - 3 $ is negative. Thus, $ f'(-1) $ is negative.

Because $ f'(-1) $ is negative, $ f(x) $ is decreasing at $ x = -1 $.