Let $g(x)=\tan (x)$ . $\newline$ Can we use the intermediate value theorem to say the equation $g(x)=0$ has a solution where $\frac{\pi}{4} \leq x \leq \frac{3 \pi}{4}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) No, since the function is not continuous on that interval. $\newline$ (B) No, since $0$ is not between $g\left(\frac{\pi}{4}\right)$ and $g\left(\frac{3 \pi}{4}\right)$ . $\newline$ (C) Yes, both conditions for using the intermediate value theorem have been met.

Full solution

Q. Let

g(x)=\tan (x)

\newline

Can we use the intermediate value theorem to say the equation

g(x)=0

has a solution where

\frac{\pi}{4} \leq x \leq \frac{3 \pi}{4}

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, since

0

is not between

g\left(\frac{\pi}{4}\right)

and

g\left(\frac{3 \pi}{4}\right)

\newline

Intermediate Value Theorem Application: 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 $f(c) = k$ . To apply this theorem to the function $g(x) = \tan(x)$ and the interval $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$ , we need to check two conditions: ( $1$ ) that $[a, b]$ $0$ is continuous on the interval, and ( $2$ ) that $[a, b]$ $1$ is between $[a, b]$ $2$ and $[a, b]$ $3$ .
Checking Continuity: First, we check the continuity of $g(x) = \tan(x)$ on the interval $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$ . The tangent function has discontinuities at odd multiples of $\frac{\pi}{2}$ , because it is undefined there (the cosine function in the denominator of $\tan(x) = \frac{\sin(x)}{\cos(x)}$ is zero at these points). Since $\frac{\pi}{2}$ is within the interval $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$ , the function $g(x)$ is not continuous on the entire interval.
Conclusion: Since $g(x) = \tan(x)$ is not continuous on the interval $[\frac{\pi}{4}, \frac{3\pi}{4}]$ , we cannot use the Intermediate Value Theorem to guarantee that there is a solution to $g(x) = 0$ on this interval. Therefore, the correct answer is A: No, since the function is not continuous on that interval.