Question

Let f(x) be a differentiable function for all values of x with f′(x) ≤ 32 and f(3) = 21, then the maximum value of f(8) is:

• A. 160
• B. 139
• C. 181
• D. 32

Hint:

The Mean Value Theorem helps in finding the maximum or minimum values of a function given certain conditions on its derivative.

Answer 1

The Correct Option is C

Using the mean value theorem, we can write:
\[f(8) - f(3) = f'(c)(8 - 3)\]
Given that \( f'(x) \leq 32 \), we get:
\[f(8) - 21 = 32 \times 5\]
\[f(8) = 21 + 160 = 181\]