Question

1. the functions f and g are continuous. the function h is given by h(x)=f(g(x)) - x. the table below gives values of the functions. explain why there must be a value c for 1 < c < 5 such that h(c)= - 2.

x	1	2	3	4	5
f(x)	0	9	7	- 3	8
g(x)	4	6	- 4	1	3

Explanation:

Step1: Recall Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function \(y = h(x)\) is continuous on a closed interval \([a,b]\), and \(k\) is a number between \(h(a)\) and \(h(b)\), then there exists at least one number \(c\) in the open interval \((a,b)\) such that \(h(c)=k\).

Step2: Calculate \(h(1)\) and \(h(5)\)

First, find \(h(1)\):
\[

$$\begin{align*} h(1)&=f(g(1)) - 1\\ g(1)&=4\\ f(g(1))&=f(4)= - 3\\ h(1)&=-3 - 1=-4 \end{align*}$$

\]
Next, find \(h(5)\):
\[

$$\begin{align*} h(5)&=f(g(5)) - 5\\ g(5)&=3\\ f(g(5))&=f(3)=7\\ h(5)&=7 - 5 = 2 \end{align*}$$

\]

Step3: Apply Intermediate - Value Theorem

We want \(h(c)=-2\). Since \(h(x)\) is continuous (because \(f\) and \(g\) are continuous, and the composition and subtraction of continuous functions is continuous), and \(-4=h(1)<-2 < h(5)=2\), by the Intermediate - Value Theorem, there must be a value \(c\) in the open interval \((1,5)\) such that \(h(c)=-2\).

Answer:

There must be a value \(c\) for \(1 < c < 5\) such that \(h(c)=-2\) because \(h(x)\) is continuous on \([1,5]\), \(h(1)=-4\), \(h(5)=2\), and \(-2\) is between \(-4\) and \(2\), so the Intermediate - Value Theorem applies.