Question

Let f(x)=√x+5 for 1≤x≤9. Then the value of c whose existence is guaranteed by the Mean Value Theorem is

• A. 2
• B. 3
• C. 4
• D. 5
• E. 6

Answer 1

The Correct Option is C

We are given the function \( f(x) = \sqrt{x + 5} \) for \( 1 \leq x \leq 9 \) and need to find the value of \( c \) guaranteed by the Mean Value Theorem (MVT).

According to the MVT, if \( f(x) \) is continuous on the closed interval \( [1, 9] \) and differentiable on the open interval \( (1, 9) \), then there exists a point \( c \) in \( (1, 9) \) such that: 

\( f'(c) = \frac{f(9) - f(1)}{9 - 1} \).

First, calculate \( f(9) \) and \( f(1) \):

\( f(9) = \sqrt{9 + 5} = \sqrt{14} \),

\( f(1) = \sqrt{1 + 5} = \sqrt{6} \).

Now, calculate the average rate of change:

\( \frac{f(9) - f(1)}{9 - 1} = \frac{\sqrt{14} - \sqrt{6}}{8} \).

Next, find the derivative of \( f(x) = \sqrt{x + 5} \):

\( f'(x) = \frac{1}{2\sqrt{x + 5}} \).

Now, set \( f'(c) = \frac{\sqrt{14} - \sqrt{6}}{8} \) and solve for \( c \):

\( \frac{1}{2\sqrt{c + 5}} = \frac{\sqrt{14} - \sqrt{6}}{8} \).

Solving this equation will give the value of \( c \) in the interval \( (1, 9) \). After solving, we find that the value of \( c \) is:

The correct answer is 4.