Question

If $f(x) = {x^2}{e^{2x}}$($x > 0$) then find the local maximum value of $f(x)$
A. $\dfrac{2}{{{e^2}}}$
B. $\dfrac{{ - 1}}{{{e^2}}}$
C. ${e^2}$
D. $\dfrac{1}{{{e^2}}}$

Answer 1

Hint:
For any function $f(x)$ to be maximum first equate $f'(x) = 0$ and if $f''(x) < 0$ then $x$ will be the local maxima so first of all let us assume that $f(x) = y$ then find $\dfrac{{dy}}{{dx}} = 0$ and find $x$.

Complete step by step solution:
Here we are given the function $f(x)$ which is equal to $f(x) = {x^2}{e^{2x}}$ where $x$ is positive and we need to find the local maximum value for the function$f(x)$. Here the local maxima means the point which means maximum with their neighbouring point.
For example: If ${x_0}$ is point of the local maxima and ${x_1}{\text{ and }}{x_2}$ are points just before and after the point ${x_0}$ respectively then we can say that
$f({x_0}) > f({x_1})$ and also that $f({x_0}) > f({x_2})$
Then it is called the local maxima.
It is determined by making the function slope to zero at some point and if the derivative of the function is negative then that means that it must be the local maximum.
So here we assume that $f(x) = y$ and we are given that $f(x) = {x^2}{e^{2x}}$ where $x > 0$
So $y = {x^2}{e^{2x}}$
Now as we know that if $y = h(x)g(x)$ then we can say that $\dfrac{{dy}}{{dx}} = h'(x)g(x) + h(x)g'(x)$
Here $h'(x){\text{ and g'}}(x)$ are the derivatives of the terms $h(x){\text{ and g(x)}}$ with respect to $x$ respectively.
So here
$\Rightarrow \dfrac{{dy}}{{dx}} = {x^2}\dfrac{d}{{dx}}{e^{2x}} + {e^{2x}}\dfrac{d}{{dx}}{x^2}$
$\Rightarrow \dfrac{{dy}}{{dx}} = {x^2}2{e^{2x}} + {e^{2x}}.2x$
Now equating it to zero we get that
$
\Rightarrow \dfrac{{dy}}{{dx}} = {x^2}2{e^{2x}} + 2x{e^{2x}} = 0 \\
\Rightarrow 2{e^{2x}}(x + 1) = 0 \\
 $
So we get that either $x = 0{\text{ or }}x = - 1$
Do for $x = 0{\text{ and }}x = - 1$ $\dfrac{{dy}}{{dx}} = 0$
Now we need to check for $\dfrac{{{d^2}y}}{{d{x^2}}}$
So for $\dfrac{{{d^2}y}}{{d{x^2}}}$ we get that
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2{x^2}{e^{2x}}.2 + 2{e^{2x}}.2x + 2x{e^{2x}}.2 + 2{e^{2x}}$
So we get that
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 4{x^2}{e^{2x}} + 8x{e^{2x}} + 2{e^{2x}}$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {e^{2x}}(4{x^2} + 8x + 2)$
Now for $x = 0$ we get that
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {e^{2x}}(4{x^2} + 8x + 2) = {e^0}(0 + 0 + 2) = 2$
Which is positive and therefore at $x = 0$ it is minimum.
Now for $x = - 1$ we get that
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {e^{2x}}(4{x^2} + 8x + 2) = {e^{ - 2}}(4 - 8 + 2) = \dfrac{{ - 2}}{{{e^2}}}$ and this value is negative
So we get the local maxima at this point.
$\Rightarrow f(x) = {x^2}{e^{2x}}$
$
\Rightarrow f(x) = {( - 1)^2}{e^{ - 2}} \\
\Rightarrow f(x) = \dfrac{1}{{{e^2}}} \\
 $

Note:
If we are given that a derivative of the function $f(x)$ is $g(x)$ that is $\dfrac{{d(f(x))}}{{dx}} = g(x)$ then the integral of $g(x)$ is $f(x)$ that is $\int {g(x)dx = f(x)} $.