Question

Suppose that f(0) = -8 and f'(x) ≤ 9 for all values of x. How large can f(4) possibly be?

Answer 1

Using the Mean Value Theorem, we can find an upper bound for the value of f(4). The maximum possible value for f(4) is 28.

Step-by-step explanation:

Let's use the Mean Value Theorem to find an upper bound for the value of f(4).

Step 1:

Apply the Mean Value Theorem to the function f(x) on the interval [0, 4]. The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in the open interval (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b].

Let's apply the theorem:

f'(c) = \frac{f(b) - f(a)}{b - a}

f'(c) = \frac{f(4) - f(0)}{4 - 0}

f'(c) = \frac{f(4) + 8}{4}

Step 2:

We are given that f'(x) \leq 9 for all values of x. Therefore, we can substitute f'(c) with 9 in the equation:

9 = \frac{f(4) + 8}{4}

f(4) + 8 = 36

f(4) = 28

So, f(4) can be at most 28.