Suppose that $|x-4|\leq1 $ EDIT: Okay I got it now.
Take original:
$-1\leq x-4\leq1$
Times by $(x+4)$ on all sides
$-(x+4)\leq x^2-16 \leq x+4$
The most $x+4$ can be is 9 so,
$|x^2-16| \leq 9$

Word for word:

Suppose that $|x-4|\leq 1$.
(a) What is the maximum possible value for $|x+4|$?
(b) Show that $|x^2-16|\leq 9$


My main trouble is with (b).
For (a) I did this:
$|x-4|\leq1$
$-1\leq x-4\leq1$
Add 8 to all sides:
$7\leq x+4\leq9$
So the greatest possible value is 9. How do I solve (b) though?
EDIT: I changed the inequality on (b)
 A: It seems to me that (b) is impossible.
$$|x-4|\le 1 \implies 3\le x\le 5$$
$$\implies 9\le x^2 \le 25$$
$$\implies -7 \le x^2-16 \le 9 $$
$$\implies 7 \le |x^2-16|\le 9 $$
A: Hint: (a) Use the rule $$|y|\le a\hspace{2mm}\longrightarrow\hspace{2mm}-a\le y\le a$$ and here for $y=x-4$.
For (b) you should note that $x^2-16=(x-4)(x+4)$ and use (a) to conclude.
Remark: the rule is intuitively true since $|y|\le a$ means the distance of $y$ from the origin is less than or equal to $a$, so the value of $y$ can't exceed the particular interval $[-a,a]$ which you can see by drawing a diagram.
A: For $(b)$ you have the wrong inequality. It should be $|x^2-19|\leq 16$ since we have been given
$$ |x-4| \leq 1 \implies 3\leq x \leq 5 \implies 9\leq x^2 \leq 25 $$
$$\implies -7  \leq x^2-16 \leq 9 \dots\,. $$
Can you conclude?
A: For b) consider $x^2-16 = (x+4)(x-4)$
A: If $|x - 4| \leq 1$, using that $$|a| - |b| \leq |a-b| \leq \delta \implies |x| \leq \delta + |a|$$
we get that $|x - 4| \leq 1 \implies |x| \leq 5$. By triangle inequality, you get:
$$|x + 4| \leq |x| + 4 \leq 5 + 4 = 9$$
For the second part, notice that: $$|x^2 - 16| = |x-4||x+4| \geq~ ????$$
(Hint: use item (a))
