Let f(x)=√x+5 for 1≤x≤9. Then the value of c whose existence is guaranteed by the Mean Value Theorem is
- 2
- 3
- 4
- 5
- 6
The Correct Option is C
Solution and Explanation
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.
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