Question

a continuous function y = f(x) is known to be negative at x = 4 and positive at x = 6. why does the equation f(x)=0 have at least one solution between x = 4 and x = 6? illustrate with a sketch.
why does the equation f(x)=0 have at least one solution between x = 4 and x = 6?
a. f(x)=0 has at least one solution between x = 4 and x = 6 because f(x) must pass through all values between f(4) and f(6), regardless of whether f is continuous
b. f(x)=0 has at least one solution between x = 4 and x = 6 because f is a continuous function on the closed interval 4, 6, and if y0 is any value between f(4) and f(6), then y0 = f(c) for some c in 4, 6.
c. f(x)=0 has at least one solution between x = 4 and x = 6 because all continuous functions have at least one zero over any nonempty closed interval

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 theorem to given problem

Here, $a = 4$, $b = 6$, and since $f(4)<0$ and $f(6)>0$, the value $k = 0$ is between $f(4)$ and $f(6)$. Since $f(x)$ is continuous on $[4,6]$, by the Intermediate - Value Theorem, there exists at least one $c\in(4,6)$ such that $f(c)=0$.

Answer:

B. $f(x) = 0$ has at least one solution between $x = 4$ and $x = 6$ because $f$ is a continuous function on the closed interval $[4,6]$, and if $y_0$ is any value between $f(4)$ and $f(6)$, then $y_0=f(c)$ for some $c$ in $[4,6]$.