how do i express the following as a piecewise function: f(x)=(|x^(2)-2x|)/(x-2)

Answer 1

The answer is:

y = f(x) = { x if x<=0 or x>2

-x if 0 < x < 2

}

Here's the proof:

First, recall the definition of absolute value, a piece-wise function in and of itself:

|z| = { z if z>=0

-z if z<0

}

Next,

x^2 - 2x = x(x-2), so x=0 is a zero, but x=2 is asymptote despite being a zero in the numerator.

The intervals are (-infinity,0), (0,2), (2, infinity);

Also there is a removable discontinuity at x=2, via factor (x-2) appearing in the fractional expression.

When x<0, x-2 < 0, but the numerator is positive and the denominator is negative.

so the function is negative.

So when x<0 the function can be written x(x-2)/(x-2) = x

When 0 < x < 2, then x-2 < 0. Then numerator is negative and so is the denominator.

the numerator can be rewritten as x(2-x)/(x-2) per definition of absolute value.

Therefore the function is merely -x

When x>2, then x-2 >0. The numerator and denominator are both positive, so

the function can be rewritten as just the identity line x, as |x-2|=x-2 per definition,

which kills the denominator outright

Answer 2

Graphing this function results in this graph:

Clearly there is a "division" at x = 2 and at x = 0. So one way to denote this as a piecewise function is to break it at those points.

So:

x for x<= 0

f(x) = { -x for 0<x<2

x for x>2