Question

In $\left(-4,4\right)$, the function $f\left(x\right)={\int }_{-10}^x\left(t^4-4\right)e^{-4t}dt$ has

• A. No extrema
• B. One extremum
• C. Two extrema
• D. Four extrema

Answer 1

The correct option is C

Two extrema

Explanation for the correct option.

Find the number of extrema the function has.

It is given that $f\left(x\right)={\int }_{-10}^x\left(t^4-4\right)e^{-4t}dt$.

A function has a critical point when at that point the value of its first derivative is zero.

Now the derivative of the function is given as:

$f\text{'}\left(x\right)=\left(x^4-4\right)e^{-4x}$

Equating $f\text{'}\left(x\right)=0$ and solving for $x$ we get:

$$\begin{gathered} \left(x^4-4\right)e^{-4x}=0 \\ \Rightarrow x^4-4=0\left[e^a\ne 0\right] \\ \Rightarrow x=\pm \sqrt{2} \end{gathered}$$

So the function $f\left(x\right)$ has two critical points at $x=\sqrt{2}$ and $x=-\sqrt{2}$.

If at the critical point the second derivative is zero, then the critical point is called the point of inflection but if it is not zero, then the critical point is an extremum of the function.

Differentiate $f\text{'}\left(x\right)=\left(x^4-4\right)e^{-4x}$ with respect to $x$.

$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& \frac{\partial }{\partial x}\left(\left(x^4-4\right)e^{-4x}\right)\\ & =& -4\left(x^4-4\right)e^{-4x}+4x^3{e}^{-4x}\end{array}$

So for $x=\sqrt{2}$ and $x=-\sqrt{2}$ the second derivative is not equal to zero.

So the points $x=\sqrt{2}$ and $x=-\sqrt{2}$ are the extrema of the function $f\left(x\right)={\int }_{-10}^x\left(t^4-4\right)e^{-4t}dt$.

In $\left(-4,4\right)$, the function $f\left(x\right)={\int }_{-10}^x\left(t^4-4\right)e^{-4t}dt$ has two extrema.

Hence, the correct option is C.