## Calculus (3rd Edition)

a. 9 is the maximum value b. $x^2 − 16 ≤ 9$ is true. See proof
Solve for x: $x-4\leq1$ and $x-4\geq-1$ $x\leq5$ and $x\geq3$ a. The maximum value can be found by using the greatest value of x, which is 5 $|5+4|=9$ b. $|x^2 − 16| ≤ 9$ solve for $x$: $x^2 − 16 ≤ 9$ and $x^2 − 16 \geq -9$ $x\leq5$ and $x\geq\sqrt{7}$, this does not contradict the previous conditions that we found for x, which is $x\leq5$ and $x\geq3$. Therefore, $x^2 − 16 ≤ 9$ is true.