Is $f (x) = \frac{1 - e^{2 x}}{2 x - 4}$ $f(x)=(1-e^(2x))/(2x-4)$ increasing or decreasing at $x = - 1$ $x=-1$ ?

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Answer 1 · 2016-01-05T16:06:16

Increasing.

Find the sign of the derivative at $x = - 1$ $x=-1$ .

To find $f'(x)$ , use the quotient rule.

$f'(x)=((2x-4)d/dx(1-e^(2x))-(1-e^(2x))d/dx(2x-4))/(2x-4)^2$

The derivative of each part:

The first requires the chain rule: $d/dx(e^u)=e^u*u'$

$d/dx(1-e^(2x))=-2e^(2x)$

$d/dx(2x-4)=2$

Thus,

$f'(x)=((2x-4)(-2e^(2x))-2(1-e^(2x)))/(2x-4)^2$

$f'(x)=(-4xe^(2x)+10e^(2x)+2)/(2x-4)^2$

Find $f'(-1)$ If it's greater than $0$ , the function is increasing at that point. If it's less than $0$ , the function is decreasing at that point.

$f'(-1)=(-4(-2)e^-2+10e^-2+2)/(-2-4)^2=(18/e^2+2)/36$

Since this is $>0$ , the function is increasing when $x=-1$ .

graph{(1-e^(2x))/(2x-4) [-5.93, 6.56, -2.596, 3.647]}

Is f(x)=(1-e^(2x))/(2x-4)f(x)=1−e2x2x−4f(x)=(1-e^(2x))/(2x-4) increasing or decreasing at x=-1x=−1x=-1?