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Found in: Page 215

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# 9–12 ■ Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers $$c$$ that satisfy the conclusion of the Mean Value Theorem.9. $$f(x) = 2{x^2} - 3x + 1$$, $$(0,2)$$

The numbers $$c = 1$$ satisfies the conclusion of Mean Value Theorem.

See the step by step solution

## Step 1: Given

The function $$f(x) = 2{x^2} - 3x + 1$$ is a polynomial.

## 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 $${f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}$$.

Or, equivalently, $$f(b) - f(a) = {f^\prime }(c)(b - a)$$.

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

1. Since the function $$f(x) = 2{x^2} - 3x + 1$$ is a polynomial function, it is continuous on$$R$$.

Therefore, the function is continu$$f(x)$$ous on the closed interval $$(0,2)$$.

2. Since the function $$f(x) = 2{x^2} - 3x + 1$$ is a polynomial function, it is differentiable everywhere.

Therefore, the function is differentiable on the open interval $$(0,2)$$.

Hence, the function $$f(x)$$ satisfies the conditions of Mean Value Theorem.

Since the number c satisfies the conditions of Mean Value Theorem, it should lie on the open interval $$(0,2)$$.

## Step 4: Differentiate the function and find the values

Find the derivative of .

\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {2{x^2} - 3x + 1} \right)\\ &= 2(2x) - 3(1) + 0\\ &= 4x - 3\end{aligned}

Replace $${\rm{x }} \to c$$ in $${f^\prime }(x)$$.

$${f^\prime }(c) = 4c - 3$$

Obtain the functional values at the end points of the given interval.

Substitute $$0{\rm{ }} \to x$$ in $$f(x)$$.

\begin{aligned}{c}f(a) &= f(0)\\ &= 2{(0)^2} - 3(0) + 1\\ &= 1\end{aligned}

Substitute $$2{\rm{ }} \to x$$ in $$f(x)$$.

\begin{aligned}{c}f(b) &= f(2)\\ &= 2{(2)^2} - 3(2) + 1\\ &= 8 - 6 + 1\\ &= 3\end{aligned}