Question

Example 5 suppose that f(0) = −8 and f '(x) ≤ 9 for all values of x. how large can f(3) possibly be? solution we are given that f is differentiable (and therefore continuous) everywhere. in particular, we can apply the mean value theorem on the interval [0, 3] . there exists a number c such that f(3) − f(0) = f '(c) − 0 so f(3) = f(0) + f '(c) = −8 + f '(c). we are given that f '(x) ≤ 9 for all x, so in particular we know that f '(c) ≤ . multiplying both sides of this inequality by 3, we have 3f '(c) ≤ , so f(3) = −8 + f '(c) ≤ −8 + = . the largest possible value for f(3) is .

Answer 1

$f'(x)$ exists and is bounded for all $x$. We're told that $f(0)=-8$. Consider the interval [0, 3]. The mean value theorem says that there is some $c\in(0,3)$ such that

$f'(c)=\dfrac{f(3)-f(0)}{3-0}$

Since $f'(x)\le9$, we have

$\dfrac{f(3)+8}3\le9\implies f(3)\le19$

so 19 is the largest possible value.