Question

If \( f(x) \) is a differentiable function, \( f'(x) \geq 5 \) for \( x \in [2,6] \), \( f(2) = 4 \) and \( f(3) = 15 \), then a possible value of \( f(6) \) is:

• A. \( 24 \)
• B. lies between 4 and 15
• C. \( 15 \leq f(6) \)
• D. \( f(6) = 5 \)

Hint:

Use the Mean Value Theorem when working with differentiable functions and inequalities on their derivatives to estimate values at endpoints.

Answer 1

The Correct Option is A

We are given that \( f'(x) \geq 5 \) for \( x \in [2,6] \), which means that the function \( f(x) \) is increasing with a slope of at least 5. Using the Mean Value Theorem for the interval \( [2, 6] \), we know that: \[ f(6) - f(2) = f'(c) \cdot (6 - 2) \] for some \( c \in (2,6) \). Since \( f'(c) \geq 5 \), we have: \[ f(6) - f(2) \geq 5 \cdot (6 - 2) = 5 \cdot 4 = 20 \] Therefore, \( f(6) \geq f(2) + 20 = 4 + 20 = 24 \). Thus, the possible value of \( f(6) \) is at least 24, and hence the correct answer is option (1).

Answer 2

To determine a possible value of \( f(6) \), we need to use the information provided about \( f \). We know that the derivative \( f'(x) \geq 5 \) for \( x \in [2,6] \). According to the Mean Value Theorem, for a differentiable function \( f \), there exists some \( c \) in the interval \([2,6]\) such that:

$$f'(c) = \frac{f(6) - f(2)}{6 - 2}.$$

Using the information \( f'(x) \geq 5 \), we have:

$$\frac{f(6) - f(2)}{6 - 2} \geq 5.$$

Rearranging gives us:

$$f(6) - f(2) \geq 20.$$

Since \( f(2) = 4 \):

$$f(6) \geq 24.$$

Therefore, a possible value of \( f(6) \) is

24.