Question

write the following absolute value function as a piecewise function.
$f(x) = \left|x^2 - 3x - 4\
ight|$
answer attempt 2 out of 3
$f(x) = \

$$\begin{cases} \\square & \\text{for } \\square \\\\ \\square & \\text{for } \\square \\end{cases}$$

$

Explanation:

Step1: Factor the quadratic

First, factor \(x^2 - 3x - 4\). We find two numbers that multiply to \(-4\) and add to \(-3\), which are \(-4\) and \(1\). So, \(x^2 - 3x - 4=(x - 4)(x + 1)\).

Step2: Find critical points

Set \(x^2 - 3x - 4 = 0\). Solving \((x - 4)(x + 1)=0\) gives \(x = 4\) or \(x=-1\). These are the critical points where the expression inside the absolute value changes sign.

Step3: Test intervals

• For \(x < -1\) or \(x > 4\), let's test \(x = 5\): \(5^2 - 3(5)-4 = 25 - 15 - 4 = 6>0\). So, \(|x^2 - 3x - 4|=x^2 - 3x - 4\) when \(x\leq - 1\) or \(x\geq 4\) (including the critical points as the expression is zero there).
• For \(-1 < x < 4\), test \(x = 0\): \(0^2 - 3(0)-4=-4 < 0\). So, \(|x^2 - 3x - 4|=-(x^2 - 3x - 4)=-x^2 + 3x + 4\) when \(-1 < x < 4\).

Answer:

$$ f(x)=
LATEXBLOCK0
$$