Question

1. (4 points) use the table below to answer the following questions.

a. where are the zero(s) located?
between ( x = )____ and ( x = )____
between ( x = )____ and ( x = )____
b. there is a relative maximum or minimum.
(circle one)
c. located near ( x = )______.

Explanation:

Step1: Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \(y = f(x)\) is continuous on a closed interval \([a,b]\) and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in the interval \((a,b)\) such that \(f(c)=k\). For a zero, \(k = 0\).
When \(x=-2\), \(f(-2)=-7\) and when \(x = - 1\), \(f(-1)=8\). Since \(0\) is between \(-7\) and \(8\), by the Intermediate Value Theorem, there is a zero between \(x=-2\) and \(x=-1\).
When \(x = 1\), \(f(1)=8\) and when \(x=2\), \(f(2)=-7\). Since \(0\) is between \(8\) and \(-7\), by the Intermediate Value Theorem, there is a zero between \(x = 1\) and \(x=2\).

Step2: Analyze the relative maximum/minimum

A function has a relative maximum if \(f(x_0)\geq f(x)\) for all \(x\) in some open interval containing \(x_0\) and a relative minimum if \(f(x_0)\leq f(x)\) for all \(x\) in some open interval containing \(x_0\).
We have \(f(-2)=-7\), \(f(-1)=8\), \(f(0)=9\), \(f(1)=8\), \(f(2)=-7\). Since \(f(-1)=8\), \(f(0)=9\), \(f(1)=8\), the function changes from increasing (\(f(-2)\lt f(-1)\lt f(0)\)) to decreasing (\(f(0)>f(1)>f(2)\)). So there is a relative maximum.

Step3: Locate the relative maximum

Since \(f(-1) = 8\), \(f(0)=9\), \(f(1)=8\), the function reaches its local maximum value at \(x = 0\) (because the function value \(f(0)=9\) is greater than the function values at its adjacent points \(x=-1\) and \(x = 1\))

Answer:

a. Between \(x=-2\) and \(x=-1\); Between \(x = 1\) and \(x=2\)
b. Relative maximum (circle "Relative maximum")
c. \(x = 0\)