Question

Find the minimum and maximum value of function$f\left( x \right) = {x^2}{e^x}$.

Answer 1

Equating first order derivative of the given function to zero to get solution, and then substituting values of solution in second derivative of function to discuss point of maxima or minima. Finally substituting values in the given function to get maximum or minimum value corresponding to point of maxima or point of minima.
Formulas used: $\dfrac{{df}}{{dx}} = 0,\,\,\,\,\dfrac{{{d^2}f}}{{d{x^2}}} < 0$ for point of maxima and $\dfrac{{{d^2}f}}{{d{x^2}}} > 0$ for point of minima.

Complete step by step solution:
Given, function is $f\left( x \right) = {x^2}{e^x}$

Calculating first derivative of $f\left( x \right)$
$\dfrac{{df}}{{dx}} = {x^2}\dfrac{{d({e^x})}}{{dx}} + {e^x}\dfrac{d}{{dx}}\left( {{x^2}} \right)$
$\Rightarrow \dfrac{{df}}{{dx}} = {x^2}\left( {{e^x}} \right) + {e^x}\left( {2x} \right)$
$\Rightarrow \dfrac{{df}}{{dx}} = {e^x}\left( {{x^2} + 2x} \right)$
Now, equating$\dfrac{{df}}{{dx}}$ is equal to zero to get solution of the equation or value of x.
$\Rightarrow {e^x}\left( {{x^2} + 2x} \right) = 0$
$\Rightarrow \left( {{x^2} + 2x} \right) = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \left( {{e^x} \ne 0} \right)$
$\Rightarrow x\left( {x + 2} \right) = 0 \\\ \Rightarrow x = 0\,\,or\,\,x + 2 = 0 \\\ \Rightarrow x = 0\,\,\,or\,\,x = - 2 \\\$

Now, calculating $\dfrac{{{d^2}f}}{{d{x^2}}}$ of the given function.
$\dfrac{{{d^2}f}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{df}}{{dx}}} \right)$
\Rightarrow \dfrac{{{d^2}f}}{{d{x^2}}} = \dfrac{d}{{dx}}\left\\{ {{e^x}\left( {{x^2} + 2} \right)} \right\\}
$\Rightarrow \dfrac{{{d^2}f}}{{d{x^2}}} = {e^x}\dfrac{d}{{dx}}\left( {{x^2} + 2x} \right) + \left( {{x^2} + 2x} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)$
$\Rightarrow \dfrac{{{d^2}f}}{{d{x^2}}} = {e^x}\left( {2x + 2} \right) + \left( {{x^2} + 2x} \right)\left( {{e^x}} \right)$
$\Rightarrow \dfrac{{{d^2}f}}{{d{x^2}}} = {e^x}\left( {{x^2} + 4x + 2} \right)$

Substituting values of $x = 0$ and $x = - 2$ obtained in step 3 in $\dfrac{{{d^2}f}}{{d{x^2}}}$ obtained in step 5 to check point of maxima or point of minima.
First substituting $x = 0$ we have,
{\left( {\dfrac{{{d^2}f}}{{d{x^2}}}} \right)_{x = 0}} = {e^{(0)}}\left\\{ {{{(0)}^2} + 4(0) + 2} \right\\}
$\Rightarrow {\left( {\dfrac{{{d^2}f}}{{d{x^2}}}} \right)_{x = 0}} = 2 > 0$
Therefore we can say that point $x = 0$ is a point of minima.

Now, substituting $x = - 2$ in $\dfrac{{{d^2}f}}{{d{x^2}}}$ we have
{\left( {\dfrac{{{d^2}f}}{{d{x^2}}}} \right)_{x = - 2}} = {e^{ - 2}}\left\\{ {{{\left( { - 2} \right)}^2} + 4\left( { - 2} \right) + 2} \right\\}
${\left( {\dfrac{{{d^2}f}}{{d{x^2}}}} \right)_{x = - 2}} = \dfrac{1}{{{e^2}}}\left( {4 - 8 + 2} \right) < 0$
Therefore we can say that point $x = - 2$ is a point of maxima.

To calculate corresponding maximum and minimum values of the given function $f\left( x \right)$we put$x = 0$ and $x = - 2$in $f\left( x \right)$ one by one.
Substituting $x = 0$ we have
$f\left( x \right) = (0){e^{(0)}} = 0$
Substituting $x = - 2$ we have
$f\left( x \right) = {( - 2)^2}{e^{( - 2)}} \\\ \Rightarrow f(x) = \dfrac{4}{{{e^2}}} \\\$

Therefore, form above we see that minimum value of given function is 0 and maximum value of the given function is$\dfrac{4}{{{e^2}}}$

Note: Maxima and minima of the function can be calculated by using differentiation method.