Question

let ( f ) be a differentiable function with ( f(-3) = 7 ) and ( f(3) = 8 ). which of the following must be true for some ( c ) in the interval ( (-3, 3) )?

a ( f(c) = 0 ), since the extreme value theorem applies.

b ( f(c) = \frac{8 + 7}{3 - (-3)} ), since the mean value theorem applies.

c ( f(c) = \frac{8 - 7}{3 - (-3)} ), since the mean value theorem applies.

d ( f(c) = 7.5 ), since the intermediate value theorem applies.

Explanation:

Step1: Recall Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists some \( c \in (a, b) \) such that \( f^{\prime}(c)=\frac{f(b)-f(a)}{b - a} \). Here, \( a=-3 \), \( b = 3 \), \( f(a)=7 \), \( f(b)=8 \).

Step2: Analyze Option A

The Extreme Value Theorem (EVT) guarantees extrema on a closed, bounded interval for continuous functions. But \( f^{\prime}(c) = 0 \) is about critical points (Fermat's theorem), and EVT doesn't imply \( f^{\prime}(c)=0 \). So A is wrong.

Step3: Analyze Option B

The numerator in the slope formula should be \( f(b)-f(a) \), not \( f(b)+f(a) \). So \( \frac{8 + 7}{3-(-3)} \) is incorrect. B is wrong.

Step4: Analyze Option C

Using MVT: \( f^{\prime}(c)=\frac{f(3)-f(-3)}{3-(-3)}=\frac{8 - 7}{3-(-3)} \). Since \( f \) is differentiable (hence continuous) on \([-3,3]\), MVT applies. So C is correct.

Step5: Analyze Option D

The Intermediate Value Theorem (IVT) applies to continuous functions for function values, not derivatives. IVT doesn't guarantee \( f^{\prime}(c)=7.5 \). So D is wrong.

Answer:

C. \( f^{\prime}(c)=\frac{8 - 7}{3-(-3)} \), since the Mean Value Theorem applies.