Question

1. no calculator what are all values of ( x ) for which the function ( f ) defined by ( f(x)=(x^{2}-3)e^{-x} ) is increasing? (a) there are no such values of ( x ). (b) ( x<-1 ) and ( x>3 ) (c) ( -3<x<1 ) (d) ( -1<x<3 ) (e) all values of ( x )

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{2}-3\), \(u^\prime=2x\), \(v = e^{-x}\), \(v^\prime=-e^{-x}\).
\(f^\prime(x)=(x^{2}-3)(-e^{-x})+2x\cdot e^{-x}=e^{-x}(-x^{2}+2x + 3)\)
Since \(e^{-x}>0\) for all \(x\), we only need to consider the sign of \(-x^{2}+2x + 3\).

Step2: Solve the inequality \(-x^{2}+2x + 3>0\)

Multiply both sides by \(- 1\) (and reverse the inequality sign) to get \(x^{2}-2x - 3<0\).
Factor the quadratic: \(x^{2}-2x - 3=(x - 3)(x + 1)<0\).
The roots of \(y=(x - 3)(x + 1)\) are \(x=-1\) and \(x = 3\).
Using a sign - chart or test intervals:
If \(x<-1\), let \(x=-2\), then \((-2 - 3)(-2 + 1)=(-5)\times(-1)=5>0\).
If \(-1If \(x>3\), let \(x = 4\), then \((4 - 3)(4 + 1)=1\times5 = 5>0\).

Answer:

D. \(-1