In the interval (-4, 4),the function f(x) = i | vedclass

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(C) Given $f(x) = \int_{-10}^x (t^4 - 4)e^{-4t} dt$. By the Fundamental Theorem of Calculus,$f'(x) = (x^4 - 4)e^{-4x}$. To find the critical points,se

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Two extrema

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(C) Given $f(x) = \int_{-10}^x (t^4 - 4)e^{-4t} dt$. By the Fundamental Theorem of Calculus,$f'(x) = (x^4 - 4)e^{-4x}$. To find the critical points,set $f'(x) = 0$: $(x^4 - 4)e^{-4x} = 0$. Since $e^{-4x} 
eq 0$ for all $x$,we have $x^4 - 4 = 0$,which implies $x^2 = 2$ or $x^2 = -2$. In the real domain,$x = \sqrt{2}$ and $x = -\sqrt{2}$. Both values $\sqrt{2} \approx 1.414$ and $-\sqrt{2} \approx -1.414$ lie within the interval $(-4, 4)$. Since $f'(x)$ changes sign at $x = \sqrt{2}$ and $x = -\sqrt{2}$,the function has two extrema in the given interval.

In the interval $(-4, 4)$,the function $f(x) = \int_{-10}^x (t^4 - 4)e^{-4t} dt$ has:

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