## Precalculus: Mathematics for Calculus, 7th Edition

Evaluate $f(x) = 2 |x-1|$: $f(-2) = 6$ $f(0) = 2$ $f(\frac{1}{2}) = 1$ $f(2) = 2$ $f(x+1) = 2x$ $f(x^2+2) = 2x^2+2$
For x = -2 $f(-2) = 2 |(-2)-1|$ ... subtract (-2)-1 $=2 |-3|$ ... take absolute value of -3 $=2(3)$ ... multiply $=6$ ______________________ For x = 0 $f(0) = 2 |(0)-1|$ ... subtract (0)-1 $=2 |-1|$ ... take absolute value of -1 $=2(1)$ ... multiply $=2$ ______________________ For x = $\frac{1}{2}$ $f(\frac{1}{2}) = 2 |\frac{1}{2}-1|$ ... subtract $\frac{1}{2}-1$ $=2 |-\frac{1}{2}|$ ... take absolute value of $-\frac{1}{2}$ $=2(\frac{1}{2})$ ... multiply $=1$ ______________________ For x = 2 $f(2) = 2 |(2)-1|$ ... subtract 2-1 $=2 |1|$ ... take absolute value of 1 $=2(1)$ ... multiply $=2$ ______________________ For x = x + 1 $f(x+1) = 2 |(x+1)-1|$ ... subtract (1)-1 $=2 |x|$ ... take absolute value of x $=2(x)$ ... multiply $=2x$ ______________________ For x = $x^2+2$ $f(x^2+2) = 2 |(x^2+2)-1|$ ... subtract (2)-1 $=2 |x^2+1|$ ... take absolute value of $x^2+1$ (value will be positive for every value of x) $=2(x^2+1)$ ... multiply $=2x^2+2$ ______________________