Consider the function f(x) = 4 \sqrt {x} + 7 on the interval {2,10}. Find the average or mean...

Question:

Consider the function {eq}f(x) = 4 \sqrt {x} + 7 {/eq} on the interval {eq}[2,10] {/eq}. Find the average or mean slope of the function on this interval.

By the Mean Value Theorem, we know there exists a c in the open interval (2,10) such that f'(c) is equal to this mean slope.

For this problem, there is only one c that works. Find it.

Average or Mean Slope:

Accroding to the Mean Value Theorem, we know that if {eq}f(x) {/eq} is a function and it is differentiable in open interval {eq}(a,b) {/eq} then there exist {eq}c {/eq} in the given open interval such that:

$$f^{'} (c) = \frac{f(b) - f(a)}{b - a} $$

The value of {eq}\frac{f(b)- f(a)}{b - a} {/eq} is called Average or mean slope of the function.

Answer and Explanation:

{eq}f(x) = 4 \sqrt{x} + 7 = 4x^{\frac{1}{2}} + 7 {/eq}

Derivative of this function is:

{eq}\begin{align*} f^{'}(x) &= \frac{d}{dx}...

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What is the Mean Value Theorem?

from Math 104: Calculus

Chapter 7 / Lesson 3
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