$xe^x$ even or odd?

Question

I have this simple function:

$$xe^x$$

Is it an even or odd function? How can I prove it?

For even functions, we have:

$$f(x)=f(-x)$$

For odd functions, we have:

$$-f(x)=f(-x)$$

I have only got the hint that it is an even function but my logic says odd.

Answer 1

Note that

$$f(x)=xe^x\neq f(-x)=-xe^{-x}$$

$$f(x)=xe^x\neq -f(-x)=xe^{-x}$$

Indeed

$$xe^x=-xe^{-x}\iff x(e^x+e^{-x})=0 \iff x=0 \quad \lor \quad e^x+e^{-x}=0 $$

$$xe^x=xe^{-x}\iff x(e^x-e^{-x})=0 \iff x=0 \quad \lor \quad e^x-e^{-x}=0 $$

since the equalities hold only for $x=0$, $f(x)$ is neither odd or even.

On the other hand,

$$g(x)=xe^{|x|}$$

is odd.

Answer 2

Note that $$f (x) = xe^x \text { and } f (-x) = -xe^{-x} $$

Note that $e^x $ isn't related to $e^{-x} $ other than the fact that they are multiplicative inverses of each other and are equal only at $x =0$. Can we relate the two functions? No. $$f (x) \neq f (-x) \text { and } f (x) \neq -f (-x) $$

Hence, $f (x) $ is neither even nor odd.

Answer 3

$$f(1)=e$$ $$f(-1)=\frac{-1}{e}$$ $$f(1) \neq f(-1)$$ $$f(1) \neq -f(-1)$$

Answer 4

Note that $xe^x \to+\infty$ as $x\to+\infty$ and $xe^x\to0$ as $x\to-\infty.$ That implies this function is neither even nor odd.

(The second limit can be shown by L'Hopital's rule applied to $\dfrac x {e^{-x}}.$ It can also be shown by more common-sense methods.)