Question

How do I write f(x)=|x-1| as a piece wise function?

Answer 1

`f(x) = |x-1| = {(x-1), x>=1`

`f(x) = |x-1| = {-(x-1), x<1`

where `x>=1 and x < 1` are referred to as Constraint Intervals

Explanation:

Given:

`color(red)(f(x) = |x-1|" "` **... Absolute Value Function

Observe the following:

If `color(green)(|x-1| = (x-1)`, then

`(x-1)` is either positive or zero.

If `color(green)(|x-1| = -(x-1)`, then

`(x-1)` is negative.

For what value of `x` will `|x-1| = (x-1)?`

`x>=1`

For what value of `x` will `|x-1| = -(x-1)?`

`x<1`

Piece-wise Function is given by different expressions on various intervals

Hence, we have

`f(x) = |x-1| = {(x-1), x>=1`

`f(x) = |x-1| = {-(x-1), x<1`

where `x>=1 and x < 1` are referred to as Constraint Intervals

Hence, we have the required Piece-wise functions

Analyze the graphs given of below:

`color(red)(f(x) = |x-1|`

Graph of the Piece-wise functions are given below:

We observe that both the graphs are identical.

Hence, our solution is verified.

Answer 2

`f(x) = |x-1| = { (1-x, x lt 1), (x-1,x ge 1) :}`

Explanation:

Consider the function:

`y= |x|`

By the definition n of the "modulus" or "absolute" value function we require that iut return the input value with a positive sign. We can write this as a piecewise function as follows:

`y = |x| = { (-x, x lt 0), (0,x=0),(x,x gt 0) :}`

We further note that we can arbitrarily incorporate the case `x=` into either of the other conditions, so let us further refine our function:

`y = |x| = { (-x, x lt 0), (x,x ge 0) :}`

Then the required function, `y=f(x)` is a translation of the function `y=|x|` to the right by 1 unit, thus:

`f(x) = |x-1| = { (-(x-1), (x-1) lt 0), ((x-1),(x-1) ge 0) :}`

graph{|x-1| [-10, 10, -5, 5]}