Question

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

Answer 1

Increasing.

Explanation:

If `f'(-1)<0`, then `f(x)` is decreasing at `x=-1`.
If `f'(-1)>0`, then `f(x)` is increasing at `x=-1`.

Find the first derivative.

Use the quotient rule.

`f'(x)=((x-2)d/dx[x^2-e^x]-(x^2-e^x)d/dx[x-2])/(x-2)^2`

Find each derivative separately.

`d/dx[x^2-e^x]=2x-e^x`

`d/dx[x-2]=1`

Plug them back in.

`f'(x)=((x-2)(2x-e^x)-(x^2-e^x))/(x-2)^2`

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

`f'(x)=(x^2-xe^x+3e^x-4x)/(x-2)^2`

Find `f'(-1)`.

`f'(-1)=((-1)^2-(-1)e^(-1)+3e^(-1)-4(-1))/((-1)-2)^2`

`f'(-1)=(5+4/e)/9`

We could determine the exact value of this number, but it's clear that it will be positive. Because of this, we know that `f(x)` is increasing when `x=-1`.