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Consider the function below. f(x)=x^2 e^-x. Find minimum and maximum value of the function.

Find the value of x for which $f$ increases rapidly.

In this question we have to find the maximum and minimum value of the given function $ f\left(x\right)=x^2 \ e^{-x}$ for $x \geq 0$. We also have to find the value of x for which the given function rapidly increases.

The basic concepts behind this question are the knowledge of derivatives and the rules such as the product rule of derivatives and the quotient rule of derivatives.

Expert Answer

(a) To find out the maximum and minimum value of a given function, we have to take its first derivative and put it equal to zero to find its critical point and then put those values into the function to have maximum and minimum values.

Given function:

\[ f\left(x\right)=x^2 e^{-x}\]

For first derivative, take derivative with respect to x on both sides:

\[f^{\prime}\left(x\right)=\frac{d}{dx}\ \left[x^2 e^{-x}\right]\]

\[f^{\prime}\left(x\right)=\frac{d}{dx}[ x^2\ ] e^{-x} + x^2\frac{d}{dx} [e^{-x}]\]

\[f^{\prime}\left(x\right)=2x e^{-x}+x^2[-e^{-x}]\]

\[f^{\prime}\left(x\right)=2x e^{-x}-x^2 e^{-x}\]

\[f^{\prime}\left(x\right) =x e^{-x}(2-x)\]

Now putting the first derivative equal to zero, we get:

\[xe^{-x}(2-x)=0\]

\[xe^{-x}=0;(2-x)=0\]

\[x =0;x=2\]

Now we will find the Minimum and Maximum values of the function.

To get the minimum value put $x=0$ in the given function:

\[f\left(x\right)=x^2e^{-x}\]

\[f\left(x\right)=(0)^2e^{0}\]

\[f\left(x\right)=0\]

To get the maximum value, put $x=2$ in the given function:

\[f\left(x\right)=x^2e^{-x}\]

\[f\left(x\right)=(2)^2e^{-2}\]

\[f\left(x\right)=0.5413\]

\[f\left(x\right)=\frac{4}{ e^{2}}\]

(b) To find the exact value of $x$ at which the given function increases rapidly, take the derivative of the first derivative with respect to $x$ on both sides again.

\[f^{\prime}\left(x\right)=2x e^{-x}- x^2 e^{-x}\]

\[f^{\prime}\left(x\right)=e^{-x}(2x- x^2)\]

\[f^{\prime \prime}\left(x\right)=\frac{d}{dx}\ \left[e^{-x}(2x- x^2 \right]\]

\[f^{\prime \prime}\left(x\right)=\frac{d}{dx}\ \left(2x- x^2 \right) e^{-x}+\frac{d}{dx}\ \left(e^{-x} \right) \left(2x- x^2 \right) \]

\[f^{\prime \prime}\left(x\right) = \left(2- 2x \right) e^{-x}+ \left(-e^{-x} \right) \left(2x- x^2\right) \]

\[f^{\prime \prime}\left(x\right) = \left(2- 2x \right) e^{-x}- e^{x} \left(2x- x^2 \right) \]

\[f^{\prime \prime}\left(x\right)=e^{-x}[\left(2- 2x \right) – \left(2x- x^2\right)]\]

\[f^{\prime \prime}\left(x\right)=e^{-x}\left(2- 2x – 2x+ x^2\right)\]

\[f^{\prime \prime}\left(x\right)=e^{-x}\left(2- 4x + x^2\right)\]

\[f^{\prime \prime}\left(x\right)=e^{-x}\left(x^2- 4x +2 \right)\]

Now putting the second derivative equal to zero, we get:

\[ f^{\prime \prime}\left(x\right) = 0 \]

\[e^{-x}\left(x^2- 4x +2 \right) =0\]

\[e^{-x}=0 ; \left(x^2- 4x +2 \right) =0\]

Solving with quadratic equation:

\[x =2+\sqrt{2}; x =2-\sqrt{2}\]

Now put these values of $x$ into the first derivative to see if the answer is a positive value or negative value.

\[ f^{\prime}\left(x\right)=e^{-x}(2x- x^2)\]

\[ f^{\prime}\left(2+\sqrt{2}\right)=e^{-(2+\sqrt{2})}[2(2+\sqrt{2})- (2+\sqrt{2})^2]\]

\[f^{\prime}\left(2+\sqrt{2}\right) = -0.16\]

\[f^{\prime}\left(2+\sqrt{2}\right) < 0\]

\[f^{\prime}\left(2-\sqrt{2}\right) = e^{-(2-\sqrt{2})}[2(2-\sqrt{2})- (2-\sqrt{2})^2]\]

\[ f^{\prime}\left(2-\sqrt{2}\right)= 0.461\]

\[ f^{\prime}\left(2+\sqrt{2}\right)> 0 \]

As the value is positive when $x=2-\sqrt{2}$, so the given function increases rapidly at this value of $x$.

Numerical Result

The minimum value of the given function $f\left(x\right)=x^2 \ e^{-x}$ is at $x=0$.

The maximum value of the given function $f\left(x\right)=x^2 \ e^{-x}$ is at $x=2$.

The value is positive when $x=2-\sqrt{2}$, so the given function increases rapidly at this value of $x$.

Example

Find maximum and minimum value for $f\left(x\right)=x \ e^{-x}$.

For first derivative, take derivative with respect to $x$ on both sides:

\[f^{\prime}\left(x\right)=\frac{d}{dx}\ \left[x  e^{-x} \right]\]

\[f^{\prime}\left(x\right)=e^{-x}+x [-e^{-x}]\]

\[f^{\prime}\left(x\right)=e^{-x}(1-x)\]

\[e^{-x}=0;(1-x)=0\]

\[x =0;x=1\]

Minimum value at $x=0$

\[ f\left(x\right)=(0)e^{0}=0\]

Maximum value at $x=1$

\[ f\left(x\right)=(1)e^{-1}= e^{-1}\]

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