Video: Finding the Value of a Function at a Point Which Makes a Discontinuity Removable

A function 𝑓(π‘₯) equals (π‘₯Β² βˆ’ 9)/(π‘₯ βˆ’ 3) for all values of π‘₯ except π‘₯ = 3. For the function to be continuous at π‘₯ = 3, which of the following must be the value of 𝑓(3)? [A] 0 [B] 3 [C] ∞ [D] 6

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Video Transcript

A function 𝑓 of π‘₯ equals π‘₯ squared minus nine over π‘₯ minus three for all values of π‘₯ except π‘₯ equals three. For the function to be continuous at π‘₯ equals three, which of the following must be the value of 𝑓 of three? a) Zero, b) three, c) ∞, or d) six.

Let’s first consider why this function 𝑓 of π‘₯ is equal to the given expression for all values of π‘₯ except when π‘₯ is equal to three. If we were to attempt to evaluate 𝑓 of three by direct substitution, we’d have that 𝑓 of three is equal to three squared minus nine over three minus three. That’s nine minus nine, which is zero, over three minus two [three], which is also zero. So, we find that 𝑓 of three is equal to zero divided by zero, which is undefined.

This is why 𝑓 of π‘₯ only equals the given expression for values of π‘₯ not equal to three because π‘₯ equals three leads to a problem. However, we want the function to be continuous at π‘₯ equals three. So, we need to find another way to work out the value of 𝑓 of three. We’re going to find an alternative but equivalent way to express 𝑓 of π‘₯. And in order to do this, we need to notice that in the numerator of 𝑓 of π‘₯, we have π‘₯ squared minus nine, which is a difference of two squares.

It can therefore be factored as π‘₯ minus three multiplied by π‘₯ plus three. And now we have an alternative way of expressing our function 𝑓 of π‘₯. But what we should notice is we have a shared factor of π‘₯ minus three in the numerator and denominator. So, this shared factor can be canceled. We’re left then with 𝑓 of π‘₯ is simply equal to π‘₯ plus three. And this will be defined for all values of π‘₯. And so, it will be defined in particular when π‘₯ equals three.

Substituting π‘₯ equals three, we find that 𝑓 of three is equal to three plus three, which is equal to six. This type of discontinuity is called a removable discontinuity. It’s a single point at which the function is undefined. There is a gap in the graph of the function at this point, but the rest of the graph is continuous. If we can fill this point in, the entire graph will be made continuous. Hence why the discontinuity is referred to as removable.

In fact, if we were to sketch the graph of 𝑦 equals 𝑓 of π‘₯ or 𝑦 equals π‘₯ plus three in its simplified form, we’d see that this is a straight line. There would be a discontinuity at the point where π‘₯ equals three. But this discontinuity could be filled in if we are to define 𝑓 of π‘₯ to be equal to six at this point. So, by first factoring the numerator and then canceling the shared factor of π‘₯ minus three, we found a simplified expression for 𝑓 of π‘₯. We were then able to substitute π‘₯ equals three to find that the value of 𝑓 of three, which makes the function continuous at π‘₯ equals three, is six.

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