Question

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

Answer 1

decreasing.

Explanation:

You need to remember these rules:

• If `f'(x)>0` for every `x` on some interval I, then `f(x)` is increasing on
  the interval.
• If `f((x)<0` for every `x` on some interval I, then `f(x)` is
  decreasing on the interval.
• If `f((x)=0` for every `x` on some interval I, then `f(x)` is constant on the interval

So `f(x)=(x-2)e^x`; to differentiate we need to use the product rule:
`d/dx(uv)=u(dv)/dx+(du)/dxv`

Applying this rule gives us:

`f'(x)=(x-2)(e^x)+(1)(e^x)`
`:. f'(x)=xe^x-2e^x+e^x`
`:. f'(x)=xe^x-e^x`
`:. f'(x)=(x-1)e^x`

When `x=-2=>f'(-2)=(-2-1)e^-2`
When `:. f'(-2)=-3e^-2`

Remember that `A^x > 0 AA x in RR => e^-2 > 0`

Hence, `f'(-2)=-3e^-2 < 0` so we can conclude that when `x=-2` then `f(x)` is decreasing .

A graph always helps to visualise and confirm these solutions:
graph{(x-2)e^x [-10.114, 7.666, -4.41, 4.48]}