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.

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Intermediate Value Theorem

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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: To apply the intermediate value theorem, we need to check two conditions: ( $1$ ) the function must be continuous on the closed interval $a, b$ , and ( $2$ ) the value we are looking for (in this case, $0$ ) must be between the function values at the endpoints of the interval.
Evaluate Function Endpoints: First, we check if the function $f(x) = \sin(x)$ is continuous on the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$ . Since sine is a continuous function everywhere on the real number line, it is also continuous on the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$ .
Calculate Function Values: Next, we evaluate the function at the endpoints of the interval. We calculate $f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right)$ and $f\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)$ .
Check Value Position: Calculating $f\left(\frac{\pi}{6}\right)$ gives us $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ .
Apply Intermediate Value Theorem: Calculating $f(\frac{\pi}{3})$ gives us $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ .
Apply Intermediate Value Theorem: Calculating $f(\frac{\pi}{3})$ gives us $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ .Now we check if $0$ is between $f(\frac{\pi}{6})$ and $f(\frac{\pi}{3})$ . Since $0 < \frac{1}{2}$ and $0 < \frac{\sqrt{3}}{2}$ , and the sine function is increasing on the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$ , $0$ is indeed between $f(\frac{\pi}{6})$ and $f(\frac{\pi}{3})$ .
Apply Intermediate Value Theorem: Calculating $f(\frac{\pi}{3})$ gives us $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ .Now we check if $0$ is between $f(\frac{\pi}{6})$ and $f(\frac{\pi}{3})$ . Since $0 < \frac{1}{2}$ and $0 < \frac{\sqrt{3}}{2}$ , and the sine function is increasing on the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$ , $0$ is indeed between $f(\frac{\pi}{6})$ and $f(\frac{\pi}{3})$ .Since both conditions for the intermediate value theorem are met, we can conclude that there is at least one value $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ $1$ in the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$ such that $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ $3$ .