Let h(x)=|x|. Can we use the mean value theorem to say the equation h^(')(x)=0 has a solution where -1 < x < 1 ? Choose 1 answer: (A) No, since the function is not differentiable on that interval. (B) No, since the average rate of change of h over the interval -1 <= x <= 1 isn't equal to 0 . (C) Yes, both conditions for using the mean value theorem have been met.

Mean Value Theorem: The Mean Value Theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in the interval (a, b) such that f'(c) is equal to the average rate of change of the function over [a, b]. We need to check if h(x) = |x| satisfies these conditions on the interval [-1, 1].Continuity Check: First, we check for continuity. The function h(x) = |x| is continuous on the entire real line, including the closed interval [-1, 1]. Therefore, the first condition of the Mean Value Theorem is satisfied.Differentiability Check: Next, we check for differentiability. The function h(x) = |x| is differentiable everywhere except at x = 0, because the absolute value function has a sharp corner at that point. Since 0 is within the open interval (-1, 1), the function h(x) is not differentiable on the entire interval (-1, 1). Therefore, the second condition of the Mean Value Theorem is not satisfied.Conclusion: Since h(x) is not differentiable on the entire interval (-1, 1), we cannot apply the Mean Value Theorem to assert that there exists a c in (-1, 1) such that h'(c) = 0. The correct answer is (A) No, since the function is not differentiable on that interval.

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Let
h(x)=|x|.
Can we use the mean value theorem to say the equation
h^(')(x)=0 has a solution where
-1 < x < 1 ?
Choose 1 answer:
(A) No, since the function is not differentiable on that interval.
(B) No, since the average rate of change of
h over the interval
-1 <= x <= 1 isn't equal to 0 .
(C) Yes, both conditions for using the mean value theorem have been met.

Let $h(x)=|x|$ . $\newline$ Can we use the mean value theorem to say the equation $h^{\prime}(x)=0$ has a solution where \( -1

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Grade 7

One-step inequalities: word problems

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Q. Let

h(x)=|x|

\newline

Can we use the mean value theorem to say the equation

h^{\prime}(x)=0

has a solution where

-1<x<1

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, since the average rate of change of

h

over the interval

-1 \leq x \leq 1

isn't equal to

0

\newline

Mean Value Theorem: The Mean Value Theorem states that if a function $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , then there exists at least one $c$ in the interval $(a, b)$ such that $f'(c)$ is equal to the average rate of change of the function over $[a, b]$ . We need to check if $h(x) = |x|$ satisfies these conditions on the interval $[-1, 1]$ .
Continuity Check: First, we check for continuity. The function $h(x) = |x|$ is continuous on the entire real line, including the closed interval $[-1, 1]$ . Therefore, the first condition of the Mean Value Theorem is satisfied.
Differentiability Check: Next, we check for differentiability. The function $h(x) = |x|$ is differentiable everywhere except at $x = 0$ , because the absolute value function has a sharp corner at that point. Since $0$ is within the open interval $(-1, 1)$ , the function $h(x)$ is not differentiable on the entire interval $(-1, 1)$ . Therefore, the second condition of the Mean Value Theorem is not satisfied.
Conclusion: Since $h(x)$ is not differentiable on the entire interval $(-1, 1)$ , we cannot apply the Mean Value Theorem to assert that there exists a $c$ in $(-1, 1)$ such that $h'(c) = 0$ . The correct answer is (A) No, since the function is not differentiable on that interval.