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

1 Answer
Jan 10, 2017

f decreases at x=-1.

Explanation:

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.

Enjoy Maths.!