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QUESTION
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\item[(i)]
Show that the function

$$f(x)=x^2e^{-x}$$

has two stationary points and determine their nature.

\item[(ii)]
Determine the points of inflection of the function $f(x)$.

\item[(iii)]
Given that $x=1$ is an approximate solution of the equation
$3x^2e^{-x}=1$ use the Newton Raphson formula ONCE to obtain a
better approximation (correct to four decimal place).

\end{description}

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ANSWER
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\item[(i)]
$$f(x)=x^2e^{-x}$$
$$\frac{df}{dx}=x^2(-e^{-x})+e^{-x}(2x)=(2x-x^2)e^{-x}$$ There is
a stationary point when $\ds\frac{df}{dx}=0$, i.e. $2x-x^2=0\
(e^{-x}\neq0)$ therefore $x(2-x)=0,$ therefore $x=0$ or $x=2$
\begin{eqnarray*}
\frac{d^2f}{dx^2}&=&(2x-x^2)(-e^{-x})+e^{-x}(2-2x)\\
&=&e^{-x}(-2x+x^2+2-2x)\\ &=&(x^2-4x+2)e^{-x}
\end{eqnarray*}
When $\ds x=0,\ \frac{d^2f}{dx^2}=2e^{-0}=2>0$ therefore $f$ has a
minimum at $x=0$\\ When $\ds x=2,\
\frac{d^2f}{dx^2}=(2^2-4(2)+2)e^{-2}=-\frac{2}{e^2}<0$ therefore
$f$ has a maximum at $x=2$

\item[(ii)]
$\ds\frac{d^2f}{dx^2}=0$ when $\ds x^2-4x+2=0,\
x=\frac{4\pm\sqrt{((-4)^2-4(1)2)}}{2}=\frac{4\pm\sqrt{8}}{2}=2\pm\sqrt{2}$

\begin{eqnarray*}
\frac{d^3f}{dx^3}&=&(x^2-4x+2)(-e^{-x}+(2x-4)e^{-x}\\
&=&e^{-x}(-x^2+4x-2+2x-4)\\
&=&e^{-x}(-x^2+6x-6)\\&=&-(x^2-6x+6)e^{-x}
\end{eqnarray*}

When $\ds x=2+\sqrt{2},\ \frac{d^3f}{dx^3}\sim
2.83e^{-(2+\sqrt{2})}\neq0$\\ When $\ds
x=2-\sqrt{2},\frac{d^3f}{dx^3}\sim-2.83e^{-(2-\sqrt{2})}\neq0$\\
Therefore there are inflection points at $x=2+\sqrt{2}$ and
$x=2-\sqrt{2}$

\item[(iii)]
$\ds 3x^2e^{-x}=1,\ g(x)=3x^2e^{-x}-1=0,\
\frac{dg}{dx}=3\frac{df}{dx}=3(2x-x^2)e^{-x}\\
x_1=1-\frac{g(1)}{g'(1)}=1-\frac{0.1036}{2.2073}=0.9531$

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