Question

a continuous function y = f(x) is known to be negative at x = 8 and positive at x = 9. why does the equation f(x)=0 have at least one solution between x = 8 and x = 9? illustrate with a sketch. why does the equation f(x)=0 have at least one solution between x = 8 and x = 9? a. f(x)=0 has at least one solution between x = 8 and x = 9 because f is a continuous function on the closed interval 8,9, and if y0 is any value between f(8) and f(9), then y0 = f(c) for some c in 8,9. b. f(x)=0 has at least one solution between x = 8 and x = 9 because all continuous functions have at least one zero over any non - empty closed interval. c. f(x)=0 has at least one solution between x = 8 and x = 9 because f(x) must pass through all values between f(8) and f(9), regardless of whether f is continuous.

Explanation:

Step1: Recall 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 $f(c)=k$.

Step2: Apply to the given problem

Here, $a = 8$, $b = 9$, $f(8)<0$ and $f(9)>0$. We want to find when $k = 0$. Since $0$ is between $f(8)$ and $f(9)$ and $f(x)$ is continuous on $[8,9]$, there exists at least one $c\in(8,9)$ such that $f(c)=0$.

Answer:

A. $f(x)=0$ has at least one solution between $x = 8$ and $x = 9$ because $f$ is a continuous function on the closed interval $[8,9]$, and if $y_0$ is any value between $f(8)$ and $f(9)$, then $y_0 = f(c)$ for some $c$ in $[8,9]$