Question

using only the values given in the table for the function, $f(x)=x^{3}-3x - 2$, what is the interval of $x$-values over which the function is decreasing? $(-4,1)$ $(-4,-1)$ $(-1,1)$ $(-1,2)$

Explanation:

Step1: Understand the concept of a decreasing function

A function \(y = f(x)\) is decreasing on an interval if for any two values \(x_1\) and \(x_2\) in the interval with \(x_1

Step2: Check each interval

• For the interval \((-4,1)\)
• When \(x=-4\), \(f(-4)=-54\); when \(x = - 3\), \(f(-3)=-20\). Since \(-54<-20\), the function is not decreasing on \((-4,1)\) as a whole.
• For the interval \((-4,-1)\)
• When \(x=-4\), \(f(-4)=-54\); when \(x=-3\), \(f(-3)=-20\). Since \(-54 < - 20\), the function is not decreasing on \((-4,-1)\) as a whole.
• For the interval \((-1,1)\)
• When \(x=-1\), \(f(-1) = 0\); when \(x=0\), \(f(0)=-2\); when \(x = 1\), \(f(1)=-4\).
• Since \(0>-2\) and \(-2>-4\) (i.e., for \(x_1=-1,x_2 = 0\) with \(-1<0\) and \(f(-1)>f(0)\) and for \(x_1=0,x_2 = 1\) with \(0<1\) and \(f(0)>f(1)\)), the function is decreasing on \((-1,1)\).
• For the interval \((-1,2)\)
• When \(x = 1\), \(f(1)=-4\); when \(x=2\), \(f(2)=0\). Since \(-4<0\), the function is not decreasing on \((-1,2)\) as a whole.

Answer:

\((-1,1)\)