Absolute value expressions as piecewise expressions?
|x^2-1|
|x^2+x-12|
|x^2+4x+4|
(If you can show me how to do one, I’ll figure out the other 2!)
Favorite Answer
x^2 + x – 12 = (x + 4)(x – 3) and equals zero if x = -4 or 3
So the pieces are: x < -4, -4 < x < 3, and x > 3 [with two of the inequalities including = so -4 and 3 aren’t left out]
Now, if x , -4, both parts are negative so the product is positive and the absolute value bars make no difference. Similarly if x > 3 both are positive and again they don’t change it. But in between (x-3) is negative while (x + 4) is positive so the absolute value changes its sign.
Answer: (with the big curly brace)
x^2 + x – 12 if x >= 3 or x <= -4 -(x^2 + x - 12) if -4 < x < 3
- Academic Writing
- Accounting
- Anthropology
- Article
- Blog
- Business
- Career
- Case Study
- Critical Thinking
- Culture
- Dissertation
- Education
- Education Questions
- Essay Tips
- Essay Writing
- Finance
- Free Essay Samples
- Free Essay Templates
- Free Essay Topics
- Health
- History
- Human Resources
- Law
- Literature
- Management
- Marketing
- Nursing
- other
- Politics
- Problem Solving
- Psychology
- Report
- Research Paper
- Review Writing
- Social Issues
- Speech Writing
- Term Paper
- Thesis Writing
- Writing Styles