Question

If f(1) = 10 and f '(x) ≥ 3 for 1 ≤ x ≤ 6, how small can f(6) possibly be?

Answer 1

Answer:

$f'( x ) = \frac{f(b) - f(a)}{b - a} \\ = \frac{f(6) -f(a)}{6 - 1} \\ 3\leqslant \frac{f(6) - 10}{5} \\ 15\leqslant f(6) - 10 \\ f(6) \geqslant 15 + 10 \\ f(6) \geqslant 25$

Answer 2

Answer:

25

Step-by-step explanation:

Our "best" case is when $f'(x) = 3$ everywhere - if it changes it means $f(x)$ will grow faster. But if $f'(x)$ is constant, it means $f(x)$ is the straight line [tex] y-10 = 3 (x-1) [tex], which passes through the point (6, 25)