Answer
a. 9 is the maximum value
b. $x^2 − 16 ≤ 9$ is true. See proof
Work Step by Step
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.
