9. the functions f and g are continuous. the function h is given by h(x)=f(g(x)) - x. the table below gives…

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

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

Answer

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.