Question

consider the function (f(x) = 4sqrt{x} + 4) on the interval (3, 6). 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 ((3, 6)) such that (f(c)) is equal to this mean slope. for this problem, there is only one (c) that works. find it.

Explanation:

Calculate function values at endpoints

We evaluate \(f(x) = 4\sqrt{x} + 4\) at the endpoints of the interval \([3, 6]\).

$$ f(3) = 4\sqrt{3} + 4 $$

$$ f(6) = 4\sqrt{6} + 4 $$

Find the average slope

The average slope of the function on \([a, b]\) is given by the difference quotient:

$$ \text{Average Slope} = \frac{f(b) - f(a)}{b - a} $$

Substituting \(a = 3\) and \(b = 6\):

$$ \text{Average Slope} = \frac{(4\sqrt{6} + 4) - (4\sqrt{3} + 4)}{6 - 3} = \frac{4\sqrt{6} - 4\sqrt{3}}{3} = \frac{4(\sqrt{6} - \sqrt{3})}{3} $$

Find the derivative of the function

We find the derivative \(f'(x)\) to apply the Mean Value Theorem.

$$ f(x) = 4x^{1/2} + 4 $$

$$ f'(x) = 4 \cdot \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x}} $$

Solve for the value of c

By the Mean Value Theorem, we set \(f'(c)\) equal to the average slope:

$$ \frac{2}{\sqrt{c}} = \frac{4(\sqrt{6} - \sqrt{3})}{3} $$

Isolating \(\sqrt{c}\):

$$ \sqrt{c} = \frac{6}{4(\sqrt{6} - \sqrt{3})} = \frac{3}{2(\sqrt{6} - \sqrt{3})} $$

Squaring both sides to find \(c\):

$$ c = \frac{9}{4(\sqrt{6} - \sqrt{3})^2} $$

Expanding the denominator:

$$ (\sqrt{6} - \sqrt{3})^2 = 6 - 2\sqrt{18} + 3 = 9 - 6\sqrt{2} $$

$$ c = \frac{9}{4(9 - 6\sqrt{2})} = \frac{9}{12(3 - 2\sqrt{2})} = \frac{3}{4(3 - 2\sqrt{2})} $$

Rationalizing the denominator:

$$ c = \frac{3(3 + 2\sqrt{2})}{4(9 - 8)} = \frac{9 + 6\sqrt{2}}{4} $$

Answer:

Question 1

The average or mean slope of the function on this interval is <blank>\(\frac{4(\sqrt{6} - \sqrt{3})}{3}\)</blank>.

Question 2

The value of \(c\) that satisfies the Mean Value Theorem is <blank>\(\frac{9 + 6\sqrt{2}}{4}\)</blank>.