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

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

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

Answer

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 (-1<x<1), let (x = 0), then ((0 + 1)(0 - 1)=1\times(-1)=-1<0). If (x>1), let (x = 2), then ((2 + 1)(2 - 1)=3\times1=3>0).

Answer:

(-1<x<1)