In the interval $(-4,4)$, how many extrema does the following function have?
$$f(x)={\int }_{-10}^x(t^4-4)e^{-4t}dt$$

Using the Fundamental Theorem of Calculus (Leibniz Rule):
$$f^{\prime }(x)=\frac{d}{dx}{\int }_{-10}^x(t^4-4)e^{-4t}dt=(x^4-4)e^{-4x}$$
For extrema, set $f^{\prime }(x)=0$:
$$(x^4-4)e^{-4x}=0$$
Since $e^{-4x}>0$ for all real $x$:
$$x^4-4=0⟹(x^2-2)(x^2+2)=0$$
$$x^2=2⟹x=\sqrt{2}\approx 1.414\text{ or }x=-\sqrt{2}\approx -1.414$$
Both values $\sqrt{2}$ and $-\sqrt{2}$ lie within the interval $(-4,4)$.Thus, the function has two critical points where the derivative changes sign, resulting in two extrema.