Let $g(x)=\cos (x)$ . $\newline$ Can we use the intermediate value theorem to say the equation $g(x)=0.8$ has a solution where $0 \leq x \leq \frac{\pi}{2}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) No, since the function is not continuous on that interval. $\newline$ (B) No, since $0$ . $8$ is not between $g(0)$ and $g\left(\frac{\pi}{2}\right)$ . $\newline$ (C) Yes, both conditions for using the intermediate value theorem have been met.

Full solution

Q. Let

g(x)=\cos (x)

\newline

Can we use the intermediate value theorem to say the equation

g(x)=0.8

has a solution where

0 \leq x \leq \frac{\pi}{2}

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, since

0

8

is not between

g(0)

and

g\left(\frac{\pi}{2}\right)

\newline

Definition of 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 $f(c) = k$ . We need to check if $g(x) = \cos(x)$ is continuous on the interval $[0, \frac{\pi}{2}]$ and if $[a, b]$ $0$ is between $[a, b]$ $1$ and $[a, b]$ $2$ .
Check Continuity of $g(x)$ : First, we check the continuity of $g(x) = \cos(x)$ on the interval $[0, \frac{\pi}{2}]$ . The cosine function is continuous everywhere on the real number line, including the interval $[0, \frac{\pi}{2}]$ .
Evaluate $g(0)$ and $g(\frac{\pi}{2})$ : Next, we evaluate $g(0)$ and $g(\frac{\pi}{2})$ . We have $g(0) = \cos(0) = 1$ and $g(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0$ . We need to determine if $0.8$ is between these two values.
Determine Position of $0.8$ : Since $0.8$ is between $0$ and $1$ , it is between $g(0)$ and $g(\frac{\pi}{2})$ . Therefore, the value $0.8$ is indeed between the values of $g(x)$ at the endpoints of the interval $[0, \frac{\pi}{2}]$ .
Conclusion of Intermediate Value Theorem: Since both conditions for the intermediate value theorem are met (the function is continuous on the interval and the value $0.8$ is between $g(0)$ and $g(\frac{\pi}{2})$ ), we can conclude that there is at least one solution to the equation $g(x) = 0.8$ in the interval $[0, \frac{\pi}{2}]$ .