Answer
Evaluate $f(x) = 2 |x-1|$:
$f(-2) = 6$
$f(0) = 2$
$f(\frac{1}{2}) = 1$
$f(2) = 2$
$f(x+1) = 2x$
$f(x^2+2) = 2x^2+2$
Work Step by Step
For x = -2
$f(-2) = 2 |(-2)-1|$ ... subtract (-2)-1
$=2 |-3|$ ... take absolute value of -3
$=2(3)$ ... multiply
$=6$
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For x = 0
$f(0) = 2 |(0)-1|$ ... subtract (0)-1
$=2 |-1|$ ... take absolute value of -1
$=2(1)$ ... multiply
$=2$
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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
$=1$
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For x = 2
$f(2) = 2 |(2)-1|$ ... subtract 2-1
$=2 |1|$ ... take absolute value of 1
$=2(1)$ ... multiply
$=2$
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For x = x + 1
$f(x+1) = 2 |(x+1)-1|$ ... subtract (1)-1
$=2 |x|$ ... take absolute value of x
$=2(x)$ ... multiply
$=2x$
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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
$=2x^2+2$
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