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

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.

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Solve quadratic inequalities

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

(C) Yes, both conditions for using the intermediate value theorem have been met.

Evaluate $g(0)$ : Evaluate $g(0)$ which is $g(x) = \cos(x)$ at $x = 0$ . $\newline$ $g(0) = \cos(0) = 1$ .
Evaluate $g(\frac{\pi}{2})$ : Evaluate $g(\frac{\pi}{2})$ which is $g(x) = \cos(x)$ at $x = \frac{\pi}{2}$ . $\newline$ $g(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0$ .
Check interval for $0.8$ : Check if $0.8$ is between $g(0)$ and $g(\frac{\pi}{2})$ . $\newline$ Since $g(0) = 1$ and $g(\frac{\pi}{2}) = 0$ , and 0 < 0.8 < 1, we can say that $0.8$ is between $g(0)$ and $g(\frac{\pi}{2})$ .
Verify continuity on interval: Verify if $g(x) = \cos(x)$ is continuous on the interval $[0, \frac{\pi}{2}]$ . $\newline$ The cosine function is continuous everywhere on the real number line, including the interval $[0, \frac{\pi}{2}]$ .
Apply intermediate value theorem: Apply the intermediate value theorem. Since $g(x)$ is continuous on $[0, \frac{\pi}{2}]$ and $0.8$ is between $g(0)$ and $g(\frac{\pi}{2})$ , by the intermediate value theorem, there must be some $c$ in $[0, \frac{\pi}{2}]$ such that $g(c) = 0.8$ .

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\newline

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\newline

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\newline

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\newline

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\newline

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\newline

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\newline

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\newline

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\newline

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\newline

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