Question

Which of the following is the maximum value of ${x^4}{e^{ - {x^2}}}$-
(a) ${e^2}$
(b) ${e^{ - 2}}$
(c) $12{e^{ - 2}}$
(d) $4{e^{ - 2}}$

Answer 1

Hint: In the above question we have to find the maximum value of ${x^4}{e^{ - {x^2}}}$. So, in this type of question we will always find the critical points by taking derivatives with respect to $x$ and then we will use a second derivative test.
The second derivative test says that if the function has critical points then if the second derivative of function is positive at that point then function has local minima at that point and on the other hand if second derivative of the function is negative at that point then the function has local maxima at that point.

Complete step-by-step answer:
The given equation in the question is
 ${x^4}{e^{ - {x^2}}}$
Let $y = {x^4}{e^{ - {x^2}}}$ -(1)
Firstly, we will find the critical points by differentiating (1) with respect to $x$ and equating it with 0.
$
  \dfrac{{dy}}{{dx}} = \dfrac{{d{x^4}{e^{ - {x^2}}}}}{{dx}} \\
    \\
 $
Now using product rule and chain rule of differentiation $\dfrac{{duv}}{{dx}} = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}$ and $\dfrac{{du\left( {v\left( x \right)} \right)}}{{dx}} = \dfrac{{du\left( {v\left( x \right)} \right)}}{{dx}}\dfrac{{dv\left( x \right)}}{{dx}}$,
$\dfrac{{dy}}{{dx}}{\text{ = }}\dfrac{{d{x^4}}}{{dx}}{e^{ - {x^2}}} + \dfrac{{d{e^{ - {x^2}}}}}{{dx}}{x^4}$
Now using $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$, we get,
$
  \dfrac{{dy}}{{dx}}{\text{ = 4}}{{\text{x}}^3}{e^{ - {x^2}}} + {e^{ - {x^2}}}\left( { - 2{x^1}} \right){x^4} \\
  {\text{ = 4}}{{\text{x}}^3}{e^{ - {x^2}}} - 2{x^5}{e^{ - {x^2}}} \\
 $
So, we get,
$
  \dfrac{{dy}}{{dx}}{\text{ = 4}}{{\text{x}}^3}{e^{ - {x^2}}} - 2{x^5}{e^{ - {x^2}}} = 0 \\
  {\text{ }} \\
 $ -(2)
Now, solving the above equation we get,
$
  {\text{ }}2{{\text{x}}^3}{e^{ - {x^2}}}\left( {2 - {{\text{x}}^2}} \right) = 0 \\
  {\text{ so, }}{{\text{x}}^3} = 0{\text{ or }}2 - {{\text{x}}^2} = 0 \\
  {\text{ x = 0 or }}{{\text{x}}^2} = 2 \\
  {\text{ x = 0 or x = }} \pm \sqrt 2 {\text{ }} \\
 $
Now the critical points are ${\text{x = 0 or x = }} \pm \sqrt 2 {\text{ }}$
Now we will use the second derivative test to find that at which critical point there is local maxima.
 The second derivative test says that if function has critical points then if second derivative of function is positive at that point then function has local minima at that point and on the other hand if second derivative of the function is negative at that point then the function has local maxima at that point.
So, now second derivative of (1) is
\[
  \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{{d^2}{x^4}{e^{ - {x^2}}}}}{{d{x^2}}} \\
    \\
 \]
Now using (2) in the above equation we get,
 \[\dfrac{{{d^2}y}}{{d{x^2}}}{\text{ = }}\dfrac{{d\left( {{\text{4}}{{\text{x}}^3}{e^{ - {x^2}}} - 2{x^5}{e^{ - {x^2}}}} \right)}}{{dx}}\]
Now again using product rule and chain rule of differentiation,$\dfrac{{duv}}{{dx}} = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}$ and $\dfrac{{du\left( {v\left( x \right)} \right)}}{{dx}} = \dfrac{{du\left( {v\left( x \right)} \right)}}{{dx}}\dfrac{{dv\left( x \right)}}{{dx}}$, we get
\[
  \dfrac{{{d^2}y}}{{d{x^2}}}{\text{ = 4}} \times {\text{3}}{{\text{x}}^2}{e^{ - {x^2}}} + {\text{4}}{{\text{x}}^3}{e^{ - {x^2}}}\left( { - 2{x^1}} \right) - \left( {2 \times 5{x^4}{e^{ - {x^2}}} + 2{x^5}{e^{ - {x^2}}}\left( { - 2{x^1}} \right)} \right){\text{ }} \\
  {\text{ = 12}}{x^2}{e^{ - {x^2}}} - 8{x^4}{e^{ - {x^2}}} - 10{x^4}{e^{ - {x^2}}} + 4{x^6}{e^{ - {x^2}}} \\
  {\text{ = 2}}{{\text{x}}^2}{e^{ - {x^2}}}\left( {6 - 4{x^2} - 5{x^2} + 2{x^4}} \right) \\
 \]
Therefore,
\[\dfrac{{{d^2}y}}{{d{x^2}}}{\text{ = 2}}{{\text{x}}^2}{e^{ - {x^2}}}\left( {2{x^4} - 9{x^2} + 6} \right)\] -(3)
Now, taking critical point $x = 0$ and putting in (3) we get,
\[
  \dfrac{{{d^2}y}}{{d{x^2}}}{\text{ = 2}}{{\text{x}}^2}{e^{ - {x^2}}}\left( {2{x^4} - 9{x^2} + 6} \right) \\
  {\text{ = 2}} \times {\text{0}}{e^{ - {0^2}}}\left( {2 \times {0^4} - 9 \times {0^2} + 6} \right) \\
  {\text{ = 0}} \\
 \]
Here at $x = 0$ the second derivative is zero so this point is called point of inflection.
Now we will take critical point $x = \pm \sqrt 2 $ and putting it in (3) we get,
\[
  \dfrac{{{d^2}y}}{{d{x^2}}}{\text{ = 2}}{{\text{x}}^2}{e^{ - {x^2}}}\left( {2{x^4} - 9{x^2} + 6} \right) \\
  {\text{ = 2}}{\left( { \pm \sqrt 2 } \right)^2}{e^{ - {{\left( { \pm \sqrt 2 } \right)}^2}}}\left( {2{{\left( { \pm \sqrt 2 } \right)}^4} - 9{{\left( { \pm \sqrt 2 } \right)}^2} + 6} \right) \\
 \]
Now we know that ${\left( { \pm \sqrt 2 } \right)^2} = 2$, we get
\[
  {\text{ = 2}}\left( 2 \right){e^{ - 2}}\left( {2\left( 4 \right) - 9\left( 2 \right) + 6} \right) \\
  {\text{ = 4}}{e^{ - 2}}\left( { - 4} \right) \\
  {\text{ = - 16}}{e^{ - 2}} \\
 \]
Now as ${e^{ - 2}}$ is always positive so,
\[\dfrac{{{d^2}y}}{{d{x^2}}}{\text{ < 0}}\]
So, the second derivative for $x = \pm \sqrt 2 $ is negative (\[{\text{ < 0}}\]) so at this point there is a local maxima (using the second derivative test).
So, by putting $x = \pm \sqrt 2 $ in (1) we will get the maximum value of ${x^4}{e^{ - {x^2}}}$,
$
  y = {x^4}{e^{ - {x^2}}} \\
  {\text{ = }}{\left( { \pm \sqrt 2 } \right)^4}{e^{ - {{\left( { \pm \sqrt 2 } \right)}^2}}} \\
  {\text{ = 4}}{e^{ - 2}} \\
 $
So, the maximum value of ${x^4}{e^{ - {x^2}}}$ is ${\text{4}}{e^{ - 2}}$.

Hence, option (d) is the correct answer.

Note: The point of inflection is the point where the second derivative of a function at that point becomes zero. An inflection point is the point on a curve at which sign of the curvature changes.
Inflection points may be stationary points but are not local maxima or minima.