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.!