Graph each function. If there is a removable discontinuity, repair the break using an appropriate piecewise-defined function.

Question

Graph each function. If there is a removable discontinuity, repair the break using an appropriate piecewise-defined function.



Answer 1

We can start here by looking at the restriction on our domain: what values can't x be? Well, the denominator of our rational function can't equal zero, so we can find our restrictions by solving the "equation" x² − 4 ≠ 0:

x² − 4 ≠ 0

x² ≠ 4

x ≠ ±2

So r(-2) and r(2) are both undefined, but how can we tell that these discontinuities are "removeable?" For this, we'll need to factor and simplify our expression first. Let's focus on the denominator first:

x³ − 2x² − 4x + 8

x²(x − 2) − 4(x − 2)

(x² − 4)(x − 2)

Our rational function is now r(x) = , and we can cancel the x² − 4 in the numerator and denominator to give us the function r(x) = x − 2.

For all values of x ≠ ±2, r(x) will look like x - 2; r(2) and r(-2) will show up on our graph as infinitesimal holes. We can "repair" these discontinuities by simply defining them to be the values 2 − 2 and -2 − 2; 0 and -4. Patching these conditions in, our piecewise function looks like

Answer 2

First notice that the denominator is x² - 4, so the function would be undefined (there would be a hole in the graph) where x² - 4 = 0. So, x² = 4 and x = -2 or 2. So there are holes in the graph at x = -2 and x = 2. Then simplify the function by factoring the numerator: x³ - 2x² -4x + 8 = x² (x - 2) - 4(x - 2) = (x² - 4)(x - 2). The function r(x) then equals (x² - 4)(x - 2) / (x² - 4). The (x² - 4) cancels out in the numerator and denominator, so the function then equals x - 2. Because of the original function, there are still holes at x = -2 and 2. Evaluate x - 2 at those points and get r(-2) = -4 and r(2) = 0.

The complete piecewise function to repair the discontinuity is: r(x) = x - 2 for x ≠ -2, 2; r(x) = -4 for x = -2; r(x) = 0 for x = 2.

Then the graph would be the line x - 2. I assume from the question it should be graphed with the holes, so there would be holes at the points (-2,-4) and (2,0).