Piecewise Functions

The function defined for x = 2 is not the same as the function defined at x<2 or x>2

The function at x< 0 has the same output value as the function at x > 0

The domain of the piecewise function is the union of all of the x-values of the different intervals

If each interval of each unique function is continous with the interval that precedes it (no gaps), then the domain is defined by the interval from the lowest value of x to the greatest value of x

If the intervals are not continuous, then the domain is the union of each unique interval.

To find the domain, look at the defined intervals

Check to see if the x values are continuous (the beginning value of one function picks up where the ending of the previous function left off)

In the example, x = -1 is not included in the right-most function, but it is included in the left-most function, so the domain is continuous and includes the interval between the smallest value of x ( - $\infty$ ) to the largest value of x (+ $\infty$ ) 

This makes the domain the set of all real numbers

In this image, the domain is not defined when -2 < x < 2 and is notcontinuous. It must be named by the union of each separate interval

{ x< -2}    $\cup$   {x > 2} 

To find the range, look at the lowest function value of all the "pieces" of the function (draw a dotted line from the lowest value of the function to the y-axis). That is the lowest boundary of your range.

Repeat for the greatest value of y. Draw a line from the graph to the y-axis. This is the upper boundary for range.

If the output (includes all intervals) has defined values for all the points between the upper boundary and lower boundary, then the function is continuous.

If the function values are not defined at all points between the upper boundary and lower boundary, then it is discontinous and must be named as the union of the different intervals of f(x)

In this example, O look at my lowest value of any f(x), and that is 1/5. The greatest value is up to, but not including 5/2

Even though there are gaps between the individual intervals, if you were to shade in the intervals on the y-axis, you would see that they overlap, making them continuous

Even though this function looks like it would not be, the range is (-5/2, 1/5] since f(x) is defined at all the points in between

Look to see where the input value falls into the defined intervals

Evaluate the input using the corresponding function rule for that interval

Since 5 is > 3, you would use the function rule f(x) = 1/2 x + 1 to evaluate the function at x = 5

Since -4 < 3, you would use the function rule f(x) = 2 |x+4| -2 to evalute the function at x = -4

Since 3 > 3, you would use the function rule f(x) = 1/2 x + 1