Video: APCALC03AB-P1A-Q09-846169571895

At which point does the function 𝑓(π‘₯) = (π‘₯Β² βˆ’ 2π‘₯ βˆ’ 3)/(π‘₯Β² + 2π‘₯ βˆ’ 15) have a removable discontinuity?

03:29

Video Transcript

At which point does the function 𝑓 of π‘₯, which is equal to π‘₯ squared minus two π‘₯ minus three over π‘₯ squared plus two π‘₯ minus 15, have a removable discontinuity?

We will begin by reminding ourselves what is meant by this term removable discontinuity. A removable discontinuity is essentially a hole in a graph. It can be repaired by filling in a single point. That is, we can make the function continuous if we redefine the function at this single point or perhaps at a finite number of points if thereβ€²s more than one removable discontinuity. This differs from a nonremovable discontinuity, which would instead be in the form of a vertical asymptote.

Letβ€²s first consider the points at which this function has any type of discontinuity. And then weβ€²ll lay to determine whether or not they are removable. Discontinuities will exist when our function 𝑓 of π‘₯ is undefined. And in this case, this will occur when the denominator of the fraction is equal to zero. We can therefore find the π‘₯-values at any discontinuities by solving the quadratic equation π‘₯ squared plus two π‘₯ minus 15 equals zero. This quadratic can be factored. Itβ€²s equivalent to π‘₯ plus five multiplied by π‘₯ minus three.

To solve, we set each factor in turn equal to zero and then solve the resulting linear equation, giving π‘₯ equals negative five and π‘₯ equals three. This tells us that our function 𝑓 of π‘₯ has discontinuities at π‘₯ equals negative five and π‘₯ equals three. But are they removable discontinuities? Well, to see this, we need to consider the numerator of our fraction. Because if either of these linear factors can be cancelled from the numerator, then the gap in our function will be able to be repaired by filling in a single point.

The function in the numerator is π‘₯ squared minus two π‘₯ minus three, which factors as π‘₯ plus one multiplied by π‘₯ minus three. The function 𝑓 of π‘₯ in factored form is therefore equal to π‘₯ plus one multiplied by π‘₯ minus three over π‘₯ plus five multiplied by π‘₯ minus three. And we see that we do indeed have a common factor of π‘₯ minus three in both the numerator and denominator, which we could cancel. This tells us that the discontinuity π‘₯ equals three is removable.

Now, 𝑓 of three itself is undefined in our original function 𝑓 of π‘₯ because both the numerator and denominator of the fraction will be zero. So 𝑓 of π‘₯ would be equal to zero over zero, undefined, which is why our function has a discontinuity. However, if we substitute π‘₯ equals three into this alternative definition of 𝑓 of π‘₯, we have three plus one over three plus five. Which is four-eighths, which simplifies to one-half. And so we see that if we were to redefine the value of 𝑓 of three to be one-half, then we would be able to fill in the missing point in our graph.

So we can conclude that our function 𝑓 of π‘₯ has a removable discontinuity at the point three, one-half. The discontinuity at negative five though is nonremovable, because we donβ€²t have a common factor of π‘₯ plus five in the numerator which can be cancelled. And so this discontinuity would correspond to a vertical asymptote of the graph.

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