Question

Write y=|x|+15 as a piecewise function.

Can anyone explain this to me, please?

Answer 1

Answer:

$y = -x + 15 \: \: \: \: \: \: \: \: x < 0 \\ y = x + 15\: \: \: \: \: x \geqslant 0$

Step-by-step explanation:

Any function under mod has a break or a point of discontinuity when the function changes it's sign. To better illustrate this, let's examine a function

$y = |x - 2|$

This function will be 0 at

$x - 2 = 0 \\ x = 2$

This implies that if the input or the value of x is greater than 2, the the function's output is positive, and if smaller than 2, then the function's output is negative.

When a function is inside the mod function or absolute value function, we only take the positive signed value. To find a working pattern of this function, we can break it into two pieces, one where the value of the function under absolute value is positive and one where the value of the function under absolute value is negative.

Let's tackle this problem logically,

When the value of the function under mod is positive, say for our above stated example, then we need not make any change in the function since positive values will be unchaged. So the piece of the function where the input is greater than 2 or

$x \geqslant 2$

Our function will be,

$y = x - 2$

Now notice that for values of input lesser than 2, the value of the function inside modulus will become negative, however we have to make sure only the positive value is outputted. What are the steps we could do? Well we could square it and then take the square root, that will make sure that the value is absolute, but, there's a much easier solution. Since we already know that all the inputs lesser than 2 will fetch us with a negative output under modulus, our easiest step is to just multiply the whole thing by -1. Let's picture how that would be like

$x < 2 \\ y = ( - 1)(x - 2) \\ y = - (x - 2) \\ y = 2 - x$

By doing this step, we're making sure that we're outputting a positive value for all inputs of x, hence retaining the properties of the modulus function.

Now to find the answer of the above function

$y = |x| + 15$

We notice that for all inputs under 0, the function inside modulus, x will be negative and vice versa, so we can break the function about the value x=0

$y = -x + 15 \: \: \: \: \: \: \: \: x < 0 \\ y = x + 15\: \: \: \: \: x \geqslant0$

That shall be your answer, hope the explanation was clear for you to understand.