Question

on which interval is the function $f(x)=-x^{3}+3x - 1$ increasing?
$x < - 1$
$-1 < x < 1$
$-2 < x < 1$
$x > 1$

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=-x^{3}+3x - 1\) is \(f^\prime(x)=-3x^{2}+3\).

Step2: Set the derivative greater than zero

We want to find where \(f^\prime(x)>0\), so \(-3x^{2}+3>0\).
Divide both sides by \(- 3\) (and reverse the inequality sign): \(x^{2}-1<0\).
Factor the left - hand side: \((x + 1)(x - 1)<0\).

Step3: Solve the inequality

The solutions of the equation \((x + 1)(x - 1)=0\) are \(x=-1\) and \(x = 1\).
Using a sign chart or test intervals:
If \(x<-1\), let \(x=-2\), then \((-2 + 1)(-2 - 1)=(-1)\times(-3)=3>0\).
If \(-1If \(x>1\), let \(x = 2\), then \((2 + 1)(2 - 1)=3\times1=3>0\).

Answer:

\(-1