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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!

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You have to solve the equation $$\frac{1}{2}e^{-x}(1-x)=0$$ since $$\frac{1}{2}e^{-x}\neq 0$$ so $$1-x=0$$

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  • $\begingroup$ I understand, just 1 - x = 0 got it! $\endgroup$ – Shaun Weinberg Feb 16 at 16:57
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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 $1-x$.

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