# The function $f:\mathbb R^+ \rightarrow \mathbb R$ defined by $f(x)=e^{x^2/2}\int_{0}^{x}e^{-t^2/2}dt$ is

I came across the following problem:
The function $f:\mathbb R^+ \rightarrow \mathbb R$ defined by
$f(x)=e^{x^2/2}\int_{0}^{x}e^{-t^2/2}dt$ is
(A)monotone increasing,
(B)monotone decreasing,
(C)constant,
(D)periodic.

I have computed $f'(x)=xe^{x^2/2}\int_{0}^{x}e^{-t^2/2}dt+1$. But from here, I can not progress. Can someone point me in the right direction? Thanks in advance for your time.

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you have already computed $f'(x)=xe^{x^2/2}\int_{0}^{x}e^{-t^2/2}dt+1$.

$x\gt0$ (given), $e^{x^2}/2\gt0$, $e^{-t^2}/2\gt0\implies \int_0^x e^{-t^2}/2dt\gt0$

All these implies $f'(x)\geq1\gt0$. Thus, $f$ is monotonically increasing

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The derivative is obviously positive, so our function is increasing.

But we don't need the derivative. Since each of $e^{x^2/2}$ and $\int_0^x e^{-t^2/2}\,dt$ is increasing and positive, their product is increasing.

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If we're only working on the positive reals, let's look at that derivative $f'(x)$:

$x$: Positive

$e^{x^2/2}$: Positive

$e^{-t^2/2}$: Positive

$1$: Positive

$f'(x)$ is always positive and $f$ is monotonically increasing

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