Finding the interval where a function is strictly increasing

Question

So I have the function
$$f(x)=xe^{-x/2}$$
And want to find the interval where it is strictly increasing. If I'm not mistaken, this would be where $f'(x)>0$.

Deriving the function, I get: $\displaystyle e^{-x/2}+\frac{-xe^{-x/2}}{2}$

Then I factor it like so: $\displaystyle e^{-x/2}(1-\frac x2)$

This means that either of these two factors cannot be $0$ since that would mean that the function isn't strictly increasing for that value of x, right?

So I created this inequality: $(1-\frac x2)>0$

Solving it gives $2>x$

Looking at the solutions the answer is $-1\le x\le 1$

I am really confused..

Answer 1

$e^x > x$ for all real $x$ and so is $e^{-x}$ Now consider only the function $(1-x/2)$. When is this bigger or smaller than zero?

Answer 2

The monotonicity theorem states that a function $f\colon [a,b]\to \mathbb R$ is continuous and derivable and $f'(x)>0$ on $(a,b)$ then the function is strictly increasing in the whole closed interval $[a,b]$. It doesn't matter if $f'(a)=0$ or $f'(b)=0$.

In general if $f'(x)\ge 0$ on an interval and there is no subinterval where $f'(x)=0$ then the function is strictly increasing.

Answer 3

Your solution seems fine, unless the domain of the function is given to be $[-1,1]$, then we can be more precise.