Question

if $f(-8) = 6$ and $f(x) \geq 2$ for $x \in -8, -4$, what is the smallest value $f(-4)$ can possibly be?
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Explanation:

Step1: Apply Mean Value Theorem

The Mean Value Theorem states that if a function \( f(x) \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a \( c \in (a, b) \) such that \( f'(c)=\frac{f(b)-f(a)}{b - a} \). Given \( f'(x)\geq2 \) for \( x\in[-8,-4] \), we know \( f \) is differentiable (hence continuous) on \([-8,-4]\). So, \( \frac{f(-4)-f(-8)}{-4 - (-8)}\geq2 \).

Step2: Substitute known value

We know \( f(-8) = 6 \) and \( -4-(-8)=4 \). Substitute these into the inequality: \( \frac{f(-4)-6}{4}\geq2 \).

Step3: Solve for \( f(-4) \)

Multiply both sides by 4: \( f(-4)-6\geq8 \). Then add 6 to both sides: \( f(-4)\geq14 \). So the smallest value \( f(-4) \) can be is 14.

Answer:

14