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Expert-verified Found in: Page 215 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # To determine the values of $$c$$ that satisfies the conclusion of the Mean Value Theorem for the interval $$(1,7)$$ using the given graph of the function.

The values of $$c$$that satisfies the conclusion of the Mean Value Theorem for the interval $$(1,7)$$ by the given graph of the function is $$\underline {c = 0.3,3,6.3}$$.

See the step by step solution

## Step 1: Given fucntion

$$f(x)$$is a continuous function on $$(1,7)$$ and $$f(x)$$ is differentiable in $$(1,7)$$.

## Step 2: Concept of Mean Value Theorem

Assume $$f$$ to be a function satisfying the following properties.

(1) $$f$$Continuous on the interval $$(a,b)$$.

(2) $$f$$ Is differentiable on the interval $$(a,b)$$.

Then there is a number $$c$$ in $$(a,b)$$ such that $${f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}$$.

## Step 3: Check the conditions of Mean Value Theorem

Here, in the given graph is $$a = 1,\;b = 7$$.

Also, from the given graph $$f(x)$$ is a continuous function on $$\left( {1,7} \right)$$ and $$f(x)$$ is differentiable in $$(1,7)$$.

Clearly, from the graph it can be seen that $$f(1) = 0.5;\;f(7) = 1.5$$.

Thus, $${f^\prime }(c)$$ is evaluated as shown below.

\begin{aligned}{c}{f^\prime }(c) &= \frac{{f(7) - f(1)}}{{7 - 1}}\\ &= \frac{{1.5 - 0.5}}{{7 - 1}}\\ &= \frac{1}{6}\\ &= 0.167\end{aligned}

It can be observed from the graph that the values of $$c = 0.3,3,6.3$$.

Therefore, the values of $$c$$ that satisfies the conclusion of the Mean Value Theorem for the interval $$(1,7)$$ by the given graph of the function is . ### Want to see more solutions like these? 