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) = x , [0, 9]

Answer 1

Answer:

$c = \displaystyle\frac{9}{4}$

Step-by-step explanation:

The following information is missing in the given question:

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

Using this we may solve the question as:

We are given the following in the question:

$f(x) = \sqrt{x}, x \in [0,9]$

We have to find the number c such that f(x) satisfies the Mean value theorem.

Mean Value theorem:

It states that if the function is differentiable in the closed interval [a,b], differentiable in the interval (a,b), then there exist c in (a,b) such that:

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

Now,

$f(x) = \sqrt{x}\\f'(x) = \frac{1}{2\sqrt{x}}$

Continuity in [0,9]

Since a polynomial function is continuous everywhere, f(x) is continuous in [0,9]

Differentiability in (0,9)

Since a polynomial function is differentiable everywhere the given function is differentiable in interval (0,9)

Then, by mean value theorem:

$f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}\\\\\frac{1}{2\sqrt{c}} = \frac{f(9) - f(0)}{9-0} = \frac{3}{9}\\\\\frac{1}{2\sqrt{c}} = \frac{1}{3}\\\\c= \frac{9}{4}$