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Let f(x) be a differentiable function in...

Let f(x) be a differentiable function in [2, 7] . If f(2) = 3 and f'(x) `lt=` 5 for all x in (2, 7), then the maximum possible value of f (x) at x=7 is

A

7

B

15

C

28

D

14

Text Solution

AI Generated Solution

The correct Answer is:
To find the maximum possible value of \( f(7) \) given the conditions, we can use the Mean Value Theorem (MVT). Here’s a step-by-step solution: ### Step 1: Understand the given information We know that: - \( f(x) \) is differentiable on the interval \([2, 7]\). - \( f(2) = 3 \). - \( f'(x) \leq 5 \) for all \( x \) in the interval \((2, 7)\). ### Step 2: Apply the Mean Value Theorem According to the Mean Value Theorem, there exists at least one point \( c \) in the interval \((2, 7)\) such that: \[ f'(c) = \frac{f(7) - f(2)}{7 - 2} \] ### Step 3: Substitute the known values Substituting the known values into the equation: \[ f'(c) = \frac{f(7) - 3}{5} \] ### Step 4: Use the condition on the derivative Since \( f'(x) \leq 5 \) for all \( x \) in \((2, 7)\), we can write: \[ \frac{f(7) - 3}{5} \leq 5 \] ### Step 5: Solve the inequality Now, multiply both sides by 5: \[ f(7) - 3 \leq 25 \] Adding 3 to both sides gives: \[ f(7) \leq 28 \] ### Step 6: Conclusion Thus, the maximum possible value of \( f(7) \) is: \[ \boxed{28} \]
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