Question

What are all the values of x for which the function f defined by f(x) = (x^2 - 3)e^-xis increasing?
(A) There are no such values of x
(B) x < -1 and x > 3
(D) -1 < x <3
(C) -3 < x < 1

Answer 1

I’m pretty sure it’s D

Answer 2

Function f(x) is increasing  for values of $x<-1$ and $x>3$ .

Therefore, option (B) is correct.

Given function is,

                        $f(x)=(x^{2} -3)e^{-x}\\\\f(x)=x^{2} e^{-x}-3e^{-x}$

Taking first derivative of above function.

              $f'(x)=2xe^{-x} -e^{-x} x^{2} +3e^{-x} \\\\f'(x)=e^{-x}(2x-x^{2} +3) \\\\f'(x)=-e^{-x}(x+1)(x-3)$

To find critical points, equate above derivative to zero.

We get,   x = -1 , x = 3

We have three interval,  (-∞ , -1) , (-1, 3) and (3, ∞)

Since, in interval (-∞ , -1)  and (3, ∞) function f(x) is positive.

Therefore, In interval  (-∞ , -1)  and (3, ∞) function is increasing.

Thus, option B is correct.

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