Question

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x)

Answer 1

Answer:

2.25

Step-by-step explanation:

The computation of the number c that satisfied is shown below:

Given that

$f(x) = \sqrt{x}$

Interval = (0,9)

According to the Rolle's mean value theorem,

If f(x) is continuous in {a,b) and it is distinct also

And, f(a) ≠ f(b) so its existance should be at least one value

i.e

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

After this,

$f(x) = \sqrt{x} \\\\ f^i(x) = \frac{1}{2}x ^{\frac{1}{2} - 1} \\\\ = \frac{1}{2}x ^{\frac{-1}{2}$

$f^i(x) = \frac{1}{{2}\sqrt{x} } = f^i(c) = \frac{1}{{2}\sqrt{c} } \\\\\a = 0, f (a) = f(o) = \sqrt{0} = 0 \\\\\ b = 9 , f (b) = f(a) = \sqrt{9} = 3$

After this,

Put the values of a and b to the above equation

$f^i(c) = \frac{f(b) - f(a)}{b - a} \\\\ \frac{1}{{2}\sqrt{c} } = \frac{3 -0}{9-0} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{3}{9} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{1}{3} \\\\ \sqrt[2]{c} = 3\\\\\sqrt{c} = \frac{3}{2} \\\\ c = \frac{9}{4}$

= 2.25