Question

Use the continuity of the absolute value function (Exercise 66 ) to determine the interval(s) on which the following functions are continuous. $$f(x)=\left|x^{2}+3 x-18\right|$$

Short answer

Question: Determine the intervals of continuity for the function \(f(x)=|\,x^2+3x-18\,|\) using the obtained intervals. Answer: The function is continuous on the intervals \((-\infty, -6)\), \((-6, 3)\), and \((3, \infty)\).

Step-by-step solution

Determine non-negative intervals for the function inside the absolute value

First, we need to find the intervals where the function inside the absolute value, \(x^2+3x-18\), is non-negative. To do this, we set it greater than or equal to zero and solve for x: $$x^{2}+3x-18\ge0$$ This is a quadratic inequality, and we can find its roots by factoring or using the quadratic formula. In this case, we can factor the quadratic: $$(x-3)(x+6)\ge0$$ Now solve when the expression equals zero: $$x=3, -6$$ These are the critical points that divide the number line into intervals. Test each interval with a test point to find where the expression is non-negative: 1. Interval \((-\infty,-6)\): Test point \(x=-7\), giving \((-7-3)(-7+6)<0\) 2. Interval \((-6,3)\): Test point \(x=0\), giving \((0-3)(0+6)>0\) 3. Interval \((3,\infty)\): Test point \(x=4\), giving \((4-3)(4+6)>0\) Therefore, the function inside the absolute value is non-negative in the intervals \((-6, 3)\) and \((3, \infty)\).

Define the piecewise function based on the intervals

The given function is an absolute value function. We will now define it piecewise based on the non-negative intervals found in the previous step: $$f(x)= \begin{cases} -\left(x^{2}+3x-18\right) & -\infty<x\leq-6 \\ x^{2}+3x-18 & -6<x\leq3 \\ x^{2}+3x-18 & 3<x<\infty \end{cases} $$

The final intervals of continuity

We know that each piece of the piecewise function is continuous everywhere: the first and third pieces are quadratic functions, while the second piece is the original inside function \(x^2+3x-18\). Since the only points of interest in this case are \(x=-6\) and \(x=3\), we only need to check for continuity at these two points: 1. At \(x=-6\): \(\lim_{x\to-6^{-}}f(x)=-\left((-6)^{2}+3(-6)-18\right)=0\) and \(\lim_{x\to-6^{+}}f(x)=(x+6)(x-3)\ge1\). Thus, \(f(x)\) is not continuous at \(x=-6\). 2. At \(x=3\): \(\lim_{x\to3^{-}}f(x)=(3)^{2}+3(3)-18=0\) and \(\lim_{x\to3^{+}}f(x)=(3-3)(3+6)>0\). Thus, \(f(x)\) is not continuous at \(x=3\). So, the final intervals of continuity for the given function are \((-\infty, -6)\) and \((-6, 3)\) and \((3, \infty)\).

Key concepts

Absolute Value Function

The absolute value function, denoted as \(|x|\), transforms any negative input into a positive output, while leaving non-negative numbers unchanged.
This unique property affects how we approach solving equations and inequalities that involve absolute values. To better understand it, let's consider its impact on the continuity of functions containing absolute values.
When dealing with absolute value functions like \(f(x) = |x^2 + 3x - 18|\), it's key to first address the expression inside the absolute value, which is \(x^2 + 3x - 18\) in this case.
**Key Points:**

• The absolute value can be thought of as a piecewise function, having two conditions:
• If the expression inside is non-negative, \(f(x)\) reflects the expression as it is.
• If the expression inside is negative, \(f(x)\) returns the negative of that expression to ensure non-negativity.

Quadratic Inequality

Quadratic inequalities are expressions that include a quadratic polynomial to which a relational operator is applied, such as \(>\), \(<\), \(\geq\), or \(\leq\).
In our exercise, we have the inequality \(x^2 + 3x - 18 \geq 0\), which helps identify where the function inside our absolute value is non-negative or zero.
To solve such inequalities, we typically:

• Calculate the roots by either factoring or using the quadratic formula. For our case, we factored it as \((x - 3)(x + 6) = 0\), which gives the roots \(x = 3\) and \(x = -6\).
• Identify the intervals on the number line that these roots create: \((-\infty, -6)\), \((-6, 3)\), and \((3, \infty)\).
• Use test points from each interval to determine where the inequality holds true. This tells us where the quadratic is positive or non-negative.

Piecewise Function

A piecewise function breaks a single function into several different expressions over distinct intervals.
For the problem in hand, once we determined where \(x^2 + 3x - 18\) is non-negative, we framed the piecewise function for \(f(x) = |x^2 + 3x - 18|\):

• For \(x \leq -6\), the function is \(- (x^2 + 3x - 18)\) as it is initially negative.
• For \(-6 < x \leq 3\), and \(x > 3\), it remains \(x^2 + 3x - 18\) since the expression inside is non-negative over these ranges.

The ability to separate a function into pieces based on certain conditions or values is crucial when analyzing and solving more complex functions.

Intervals of Continuity

In mathematics, continuity of a function means there are no interruptions, jumps, or breaks in its graph.
To determine the intervals of continuity for a piecewise function, we must ensure that each "piece" of the function is continuous and check the points where these pieces meet.
For \(f(x) = |x^2 + 3x - 18|\), each segment is continuous by itself because they are quadratic functions.
However, at \(x = -6\) and \(x = 3\), transitions happen where the absolute value affects the continuity:

• At \(x = -6\): The function transitions from \(- (x^2 + 3x - 18)\) to \(x^2 + 3x - 18\), creating a potential discontinuity.
• At \(x = 3\): Similar transition occurs which introduces discontinuity due to the absolute nature affecting negatives turning to positives and vice versa.

Ultimately, the function is continuous in the intervals \((-\infty, -6)\) and \((3, \infty)\) as those segments do not include the endpoints where jumps can occur.