Question

If $y = f(x)$ is continuous on $[0, 6]$, differentiable on $(0, 6)$, $f(0) = -2$ and $f(6) = 16$, then at some point between $x = 0$ and $x = 6$, $f'(x)$ must be equal to:

• A. $-18$
• B. $-3$
• C. $3$
• D. $14$
• E. $18$

Hint:

The Mean Value Theorem essentially says that the instantaneous slope (derivative) must equal the average slope (secant line) at some point in the interval.

Answer 1

The Correct Option is C

Concept: This problem is a direct application of the Mean Value Theorem (MVT). It states that if a function is continuous and differentiable on an interval $[a, b]$, there exists at least one point $c$ in $(a, b)$ such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Step 1: Identify the interval and function values.
The interval is $[0, 6]$, so $a = 0$ and $b = 6$. We are given $f(0) = -2$ and $f(6) = 16$.

Step 2: Apply the MVT formula.
\[ f'(c) = \frac{f(6) - f(0)}{6 - 0} \] \[ f'(c) = \frac{16 - (-2)}{6} \]

Step 3: Calculate the result.
\[ f'(c) = \frac{16 + 2}{6} = \frac{18}{6} = 3 \] Thus, $f'(x)$ must be equal to 3 at some point.