Video: Finding the Unknown in a Piecewise-Defined Function That Makes It Continuous at a Point

Find the value of π‘˜ which makes 𝑓 continuous at π‘₯ = 3, given 𝑓(π‘₯) = (π‘₯⁻¹ βˆ’ 3⁻¹)/(π‘₯Β² βˆ’ 3Β²) if π‘₯ β‰  3 and 𝑓(π‘₯) = π‘˜ if π‘₯ = 3.

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Video Transcript

Find the value of π‘˜ which makes 𝑓 continuous at π‘₯ equals three. Given 𝑓of π‘₯ equals π‘₯ to the minus one minus three to the minus one over π‘₯ squared minus three squared, if π‘₯ is not equal to three and π‘˜, if π‘₯ equals three.

By definition, the function 𝑓 is continuous at a number 𝑐, if the limit of 𝑓of π‘₯ as π‘₯ approaches 𝑐 is equal to 𝑓 of 𝑐. We want to make our function 𝑓 continuous at π‘₯ equals three. So with the function 𝑓 defined as above, we need the limit of 𝑓 of π‘₯, as π‘₯ approaches three, to be 𝑓 of three. We can just read off the value of 𝑓 of three from the definition of the function. 𝑓 of π‘₯ is π‘˜ if π‘₯ is three. So this 𝑓 of three is π‘˜. Swapping the sides of the equation, we see that the value of π‘˜ which makes 𝑓 continuous at π‘₯ equals three, is the limit of 𝑓 of π‘₯ as π‘₯ approaches three.

So our task becomes to find the value of this limit. Note that if this limit doesn’t exist for some reason, then there’s not much we can do. There’s no value of π‘˜ which will make 𝑓 continuous. And so, really, we’ve just got a hope that this limit does exist. Let’s find this limit then. The limit of the function, as π‘₯ approaches three, doesn’t depend on the value of the function at π‘₯ equals three, just the values near π‘₯ equals three. So we can replace 𝑓 of π‘₯ inside this limit by the rule for 𝑓 of π‘₯ when π‘₯ is not equal to three.

This isn’t quite written in the form of a rational function. Notice the negative exponent in the power of π‘₯ in the numerator which means that the numerator is not a polynomial. However, we could still hope that this function was continuous. And that we could evaluate the limit here, using direct substitution. Unfortunately, substituting three for π‘₯ gives zero over zero, the indeterminate form. And so, it’s not that simple. We’re going to have to do some algebra.

The first thing we can do is to turn the powers with the exponents negative one into fractions. So π‘₯ to the negative one becomes one over π‘₯. And three to the negative one becomes one over three, or a third. And in the same step, we can factor the denominator which we recognize as a difference of two squares. We have fractions in our fractions. And we can get rid of them by multiplying both numerator and denominator by three π‘₯. The denominator is easy because it’s factored. We just stick the three π‘₯ in front with the numerator we have, three π‘₯ times one over π‘₯, which is three, minus three π‘₯ times a third, which is π‘₯.

Now, if we pull out a factor of negative one from the numerator, we get negative one times π‘₯ minus three. And writing the numerator in this way allows us to see a common factor which we cancel. Having cancelled the common factor of π‘₯ minus three in the numerator and denominator, we’re hopeful that directly substituting in π‘₯ equals three will give us something which is defined, and not the indeterminate form zero over zero.

Let’s try it. We substitute three for π‘₯. Doing so, we get negative one over three times three times three plus three, which is negative one over 54. If we define the function 𝑔 by the rule 𝑔 of π‘₯ equals π‘₯ to the negative one minus three to the negative one over π‘₯ squared minus three squared, then 𝑔 is not continuous at π‘₯ equals three. This is because the function 𝑔 is not defined for an input of three. Trying to evaluate 𝑔 of three using the rule gives the indeterminate form zero over zero.

But we’ve seen that we can define the function 𝑓, where 𝑓 of π‘₯ is just 𝑔 of π‘₯, if π‘₯ is not equal to three and 𝑓 of π‘₯ is negative one over 54, if π‘₯ equals three. And this function is continuous at π‘₯ equals three. By defining 𝑓 of π‘₯ to be negative one over 54 when π‘₯ is three, we managed to patch up the discontinuity in 𝑔. When we can do this, we say that the discontinuity is removable. And so we see that 𝑔 has a removable discontinuity at π‘₯ equals three.

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