Problem

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 $k$.

Solution

Step 1 :\(f(x) = 3|x-4| - 5\)

Step 2 :\(f(x) = 3(x-4) - 5\) for \(x \geq 4\)

Step 3 :\(f(x) = 3(4-x) - 5\) for \(x < 4\)

Step 4 :\(f(4) = 3(4-4) - 5 = -5\)

Step 5 :\(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} (3(x-4) - 5) = \infty\)

Step 6 :\(\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (3(4-x) - 5) = \infty\)

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 \(x \geq 4\)

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 \(x \geq 4\)

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

Step 14 :\(\frac{10}{3}x - 7 = k\) for \(x \geq 4\)

Step 15 :\(\frac{8}{3}x + 7 = k\) for \(x < 4\)

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

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