Question

Let the function f be continuous and differentiable for all x. Suppose you are given that f(−1)=−3, and that f'(x) for all values of x. Use the Mean Value Theorem to determine the largest possible value of f(5).

Answer 1

Answer:

Step-by-step explanation:

By MVT,

$f'(c) = \frac{f(5)-f(-1)}{5-1}$

$4f'(c)= f(5) + 3\rightarrow f(5) = 4f'(c) - 3$

Moreover, $f'(x)\leq 4~~~~~~ \forall x,$

$f(5) \leq 16 - 3 =13$

Hence the largest possible value of f(5) is 13