\documentclass[a4paper,12pt]{article}
\setlength\oddsidemargin{0pt} \setlength\evensidemargin{0pt}
\setlength\topmargin{0pt}
\begin{document}
\parindent=0pt
{\bf Exam Question

Topic: DiffInt}

Show that $f(x)$ is an odd function, where
$$f(x)=\int_x^{2x}\exp\left(-t^2\right)\, dt.$$ By differentiating
the integral find the turning points of $f(x)$ and identify their
type.

 \vspace{0.5in}

{\bf Solution}

Writing down the formula for $f(-x)$ and substituting $t=-u$ gives
$$f(-x)=\int_{-x}^{-2x}\exp(-t^2)\, dt=\int_x^{2x}\exp(-u^2)\,
(-du)=-f(x).$$
\begin{eqnarray*}
f'(x)&=&2\exp(-4x^2)-\exp(-x^2)\\
&=&2\exp(-x^2)\left(\exp(-3x^2-\frac{1}{2}\right)\\ &=&0\ \
\mathrm{iff}\ \ \exp(-3x^2)=\frac{1}{2}\ \ \mathrm{i.e.}\ \
x^2=\frac{\ln2}{3}.
\end{eqnarray*}
\begin{eqnarray*}
\mathrm{For}\ \  x>0,\ f'(x)&>&0\ \  \mathrm{if}\ \
x<\sqrt{\frac{\ln2}{3}}.\\ f'(x)&<&0 \ \  \mathrm{if}\ \
x>\sqrt{\frac{\ln2}{3}}
\end{eqnarray*}
So $f$ has a maximum at $\displaystyle x=\sqrt{\frac{\ln2}{3}}$
and hence, since $f$ is odd, a minimum at $\displaystyle
x=-\sqrt{\frac{\ln2}{3}}$

\end{document}