Question

If the function $f(x) = {x^2}{e^{ - 2x}},x > 0$. Then, the maximum value of $f(x)$ is
A) $\dfrac{1}{e}$
B) $\dfrac{1}{{2e}}$
C) $\dfrac{1}{{{e^2}}}$
D) $\dfrac{4}{{{e^4}}}$

Answer 1

Hint: To solve this question, we will use the basic formula of differentiation i.e. Product rule. The product rule is a rule of differentiating functions when one function is multiplied by another. The formula for product rule of differentiation is
$\dfrac{{duv}}{{dx}} = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}$

Complete step by step answer:
As per the question we have the function
$f(x) = {x^2}{e^{ - 2x}},x > 0$.
Here let us assume that
$u = {x^2},v = {e^{ - 2x}}$
Now we will first differentiate the first function i.e.
$u = {x^2}$
We know that derivate of the form ${x^n}$ is
 $n{x^{n - 1}}$, where $n$ is the exponential power.
So we can write the derivative of ${x^2}$ as
$2{x^{2 - 1}} = 2x$
We will now solve the second function i.e.
$v = {e^{ - 2x}}$
We know the differentiation of ${e^{ - 2x}}$ is
${e^{ - 2x}}$
We will now calculate the differentiation of exponential function i.e.
$ - 2x$
We know that the derivative of $cx$, where $c$ is a constant is given by the same constant.
So the differentiation of exponential function i.e. $ - 2x$ is
$ - 2$
By putting all the values in the formula we have:
$\left( {{e^{ - 2x}}} \right)(2x) + {x^2}\left( { - 2{e^{ - 2x}}} \right)$
We will take the common factor out and it gives:
$\left( {2{e^{ - 2x}}} \right)(x - {x^2})$
We have to find the maximum value, so we will equate this to zero i.e.
$\left( {2{e^{ - 2x}}} \right)(x - {x^2}) = 0$
Now we know that exponential function can never be equal to zero, so we are left with
$x - {x^2} = 0$
Here we will again take the common factor out i.e.
$x(1 - x) = 0$
From these, we get the value $x = 0$
Or,
 $x - 1 = 0 \Rightarrow x = 1$
But we have been given in the question that $x > 0$, it means that
$x = 0$is invalid.
So the correct value is
$x = 1$
Now we will put this value of $x$ in the function and we have
${1^2}{e^{ - 2 \times 1}}$
It gives us value
${e^{ - 2}}$
We can write the above expression also as
$\dfrac{1}{{{e^2}}}$
Hence the correct option is option(C) $\dfrac{1}{{{e^2}}}$.

Note:
We should always remember the rules of differentiation such as the quotient rule. We know that the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. So if we have the function:
$f(x) = \dfrac{u}{v}$ , then the formula of quotient rule says that,
$f'(x) = \dfrac{{u'v - v'u}}{{{v^2}}}$ .