Math / Calculus

QuestionIf $f(3)=9$ and $f^{\prime}(x) \geq 2$ for $3 \leq x \leq 8$ , how small can $f(8)$ possibly be? $f(8) \geq$ $\square$

Studdy Solution

STEP 1

1. We are given that $f(3) = 9$ .
2. The derivative $f'(x) \geq 2$ for $3 \leq x \leq 8$ .
3. We need to find the minimum possible value of $f(8)$ .

STEP 2

1. Understand the relationship between the derivative and the function.
2. Use the Mean Value Theorem to relate the derivative to changes in the function.
3. Calculate the minimum possible value of $f(8)$ .

STEP 3

The derivative $f'(x) \geq 2$ implies that the function $f(x)$ is increasing at a rate of at least 2 units per unit increase in $x$ over the interval $[3, 8]$ .

STEP 4

Apply the Mean Value Theorem, which states that there exists some $c$ in $(3, 8)$ such that:
$f'(c) = \frac{f(8) - f(3)}{8 - 3}$
Given $f'(x) \geq 2$ for all $x$ in $[3, 8]$ , we can use this inequality:
$2 \leq \frac{f(8) - f(3)}{5}$

STEP 5

Solve the inequality for $f(8)$ :
$2 \leq \frac{f(8) - 9}{5}$
Multiply both sides by 5 to clear the fraction:
$10 \leq f(8) - 9$
Add 9 to both sides to solve for $f(8)$ :
$f(8) \geq 19$
The minimum possible value of $f(8)$ is $\boxed{19}$ .

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