# Use sing pattern $f'(x)$ to determine where $x$ rises and falls for $f(x) = \frac{(xe^{-x})}{2}$

Use sing pattern $$f'(x)$$ to determine where $$x$$ rises and falls for $$f(x) = \frac{(xe^{-x})}{2}$$

So worked out derivative which is:

$$f'(x) = e^{-x}(1 – x)/2$$

Need it to equal zero: $$0 = e^{-x}(1 – x)/2$$

But now i'm a little stuck, how do i work out $$x$$'s since $$e^{-x}$$ cannot equal to zero? I considered to convert to ln but that does not work. Thanks!

You have to solve the equation $$\frac{1}{2}e^{-x}(1-x)=0$$ since $$\frac{1}{2}e^{-x}\neq 0$$ so $$1-x=0$$
If a product5 of some number of things, three in this case, is zero then at least one of the things is zero. Here, the three things are $$\frac{1}{2}$$, $$\mathrm{e}^{-x}$$, and $$1-x$$. The first two are never zero, so the only zeros of the product are the zeroes of $$1-x$$.