EXAM REVISION 2
1. The diagram shows a sketch of part of the graph $y=f(x)$ where $f(x)=3|x-4|-5$
Figure 2
a State the range of $f$.
$b$ Given that $\mathrm{f}(x)=-\frac{1}{3}x+k$, where $k$ is a constant has two distinct roots, state possible values of $k$.

Step 1 ：$f(x)=3|x-4|-5$

Step 2 ：$f(x)=3(x-4)-5$ for $x\ge 4$

Step 3 ：$f(x)=3(4-x)-5$ for $x<4$

Step 4 ：$f(4)=3(4-4)-5=-5$

Step 5 ：$\underset{x\to \mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}(3(x-4)-5)=\mathrm{\infty }$

Step 6 ：$\underset{x\to -\mathrm{\infty }}{lim}f(x)=\underset{x\to -\mathrm{\infty }}{lim}(3(4-x)-5)=\mathrm{\infty }$

Step 7 ：\boxed{(-\infty, -5] \cup (-5, \infty)}\)

Step 8 ：$f(x)=-\frac{1}{3}x+k$

Step 9 ：$3|x-4|-5=-\frac{1}{3}x+k$

Step 10 ：$3(x-4)-5=-\frac{1}{3}x+k$ for $x\ge 4$

Step 11 ：$3(4-x)-5=-\frac{1}{3}x+k$ for $x<4$

Step 12 ：$3x-12-5=-\frac{1}{3}x+k$ for $x\ge 4$

Step 13 ：$12-3x-5=-\frac{1}{3}x+k$ for $x<4$

Step 14 ：$\frac{10}{3}x-7=k$ for $x\ge 4$

Step 15 ：$\frac{8}{3}x+7=k$ for $x<4$

Step 16 ：$k\in \boxed{{\displaystyle (-\mathrm{\infty },-7)\cup (7,\mathrm{\infty })}}$