Chapter 4 - Section 4.2 - The Mean Value Theorem - 4.2 Exercises - Page 292: 25

$ f(4)=16$

$f'(x) = \frac{f(4) - 10}{4-1}$ $f'(x) = \frac{f(4) - 10}{3}$ Multiply by $3$ on both sides to cancel the denominator. $3f'(x) = f(4) - 10$ $3f'(x) + 10 = f(4)$ Given: $f'(x) = 2$ $3(2) + 10 = f(4)$ $6 + 10 = f(4)$ $16 = f(4)$