Let $f(x)=\sin (x)$ . $\newline$ Can we use the intermediate value theorem to say the equation $f(x)=0$ has a solution where $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) No, since the function is not continuous on that interval. $\newline$ (B) No, since $0$ is not between $f\left(\frac{\pi}{6}\right)$ and $f\left(\frac{\pi}{3}\right)$ . $\newline$ (C) Yes, both conditions for using the intermediate value theorem have been met.

Full solution

Q. Let

f(x)=\sin (x)

\newline

Can we use the intermediate value theorem to say the equation

f(x)=0

has a solution where

\frac{\pi}{6} \leq x \leq \frac{\pi}{3}

\newline

Choose

1

answer:

\newline

(A) No, since the function is not continuous on that interval.

\newline

(B) No, since

0

is not between

f\left(\frac{\pi}{6}\right)

and

f\left(\frac{\pi}{3}\right)

\newline

Check Continuity: First, we need to check if the function $f(x) = \sin(x)$ is continuous on the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$ . The sine function is known to be continuous everywhere on the real number line.
Evaluate Endpoints: Next, we evaluate the function at the endpoints of the interval. We need to find $f\left(\frac{\pi}{6}\right)$ and $f\left(\frac{\pi}{3}\right)$ to determine if $0$ is between these two values. $\newline$ $f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ $\newline$ $f\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ $\newline$ Since $0$ is between $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ , we can say that $0$ is between $f\left(\frac{\pi}{6}\right)$ and $f\left(\frac{\pi}{3}\right)$ .
Apply Intermediate Value Theorem: The intermediate value theorem states that if a function $f$ is continuous on a closed interval $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$ , then there exists at least one number $c$ in the interval $[a, b]$ such that $f(c) = k$ . $\newline$ In our case, since $f(x) = \sin(x)$ is continuous on $[\frac{\pi}{6}, \frac{\pi}{3}]$ and $[a, b]$ $0$ is between $[a, b]$ $1$ and $[a, b]$ $2$ , there must be a value $c$ in $[\frac{\pi}{6}, \frac{\pi}{3}]$ such that $[a, b]$ $5$ .