SOLUTION: Evaluate the piecewise defined function at the indicated values. f(x) = {4x if x<-4 { x+2 If -4<_ x <_ 5 { (x-5)^2 if x > 5 f(-5)= f(-4)

Algebra ->  Rational-functions -> SOLUTION: Evaluate the piecewise defined function at the indicated values. f(x) = {4x if x<-4 { x+2 If -4<_ x <_ 5 { (x-5)^2 if x > 5 f(-5)= f(-4)      Log On


   

Question 1200349: Evaluate the piecewise defined function at the indicated values.
f(x) = {4x if x<-4
{ x+2 If -4<_ x <_ 5
{ (x-5)^2 if x > 5
f(-5)=
f(-4)=
f(4)=
f(5)=
f(8)=
Answer by math_tutor2020(3817) About Me  (Show Source):
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Answers:
f(-5)= -20
f(-4)= -2
f(4)= 6
f(5)= 7
f(8)= 9


Explanation:

Piecewise functions are a Frankenstein of sorts because they have multiple functions glued together.

Another way to look at it is that the piecewise function has split identities.

The input x will determine which piece to use.
  • If x+%3C+-4, then f%28x%29+=+4x
  • If -4+%3C=+x+%3C=+5, then f%28x%29+=+x%2B2
  • If x+%3E+5, then f%28x%29+=+%28x-5%29%5E2
If the input is x = -5, then we go for the interval x+%3C+-4. Draw out a number line to see why this is the case.
This means we pick the first piece.
f%28x%29+=+4x
f%28-5%29+=+4%28-5%29
f%28-5%29+=+-20


If the input is x = -4, then we go for the interval -4+%3C=+x+%3C=+5
This time we go for the 2nd piece.
f%28x%29+=+x%2B2
f%28-4%29+=+-4%2B2
f%28-4%29+=+-2

We stick with the 2nd piece for the input x = 4, since it is also in the interval -4+%3C=+x+%3C=+5
f%28x%29+=+x%2B2
f%284%29+=+4%2B2
f%284%29+=+6

Same goes for x = 5
f%28x%29+=+x%2B2
f%285%29+=+5%2B2
f%285%29+=+7

The last input x = 8 will involve the third piece, since we're now in the interval x+%3E+5
f%28x%29+=+%28x-5%29%5E2
f%288%29+=+%288-5%29%5E2
f%288%29+=+%283%29%5E2
f%288%29+=+9


This is what the piecewise graph looks like

The first piece 4x is in red. It is graphed only when x+%3C+-4
The second piece x+2 is in green. It is graphed only when -4+%3C=+x+%3C=+5
The third piece (x-5)^2 is in blue. It is graphed only when x+%3E+5

There are open holes at (-4, -16) and (5, 0)
The open holes are not part of the graph. Think of them as potholes in the road.
There are closed filled in endpoints at (-4, -2) and (5, 7)

I recommend either Desmos or GeoGebra as two graphing options.
Here's the link to the interactive Desmos graph
https://www.desmos.com/calculator/jkfomczjuz
This will allow you to zoom in/out, move the window around, etc.