Video: Determining the Value of a Constant That Makes a Piecewise Function Continuous

Consider the function 𝑓(π‘₯) = 2π‘₯Β³ βˆ’ 3π‘₯ + 10 for π‘₯ < 2, 𝑓(π‘₯) = π‘Žπ‘₯Β³ + 7π‘₯ βˆ’ 10 for π‘₯ β‰₯ 2. For what value of π‘Ž will the function be continuous?

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Video Transcript

Consider the function 𝑓 of π‘₯ equals two π‘₯ cubed minus three π‘₯ plus 10 for π‘₯ values less than two and π‘Žπ‘₯ cubed plus seven π‘₯ minus 10 for π‘₯ values greater than or equal to two. For what value of π‘Ž will the function be continuous?

We can see that the function 𝑓 of π‘₯ is a piecewise function. It has been defined differently either side of π‘₯ equals two. Now, each piece of this function is a polynomial, so the pieces are individually continuous. To be continuous throughout its domain, which for this function is the entire set of real numbers, we need to consider what happens at the point where the definition changes. That’s the point where π‘₯ is equal to two.

And what we require for this function to be continuous is that the left-hand limit of the function as we approach two from below is equal to the value of the function itself when π‘₯ equals two. And this must also be equal to the right-hand limited of the function as we approach two from above. We can work out each of these values using direct substitution.

For the left-hand limit of the function, first of all then, we’re using the first part of the function’s definition, the part that is valid for π‘₯ values less than two. Substituting two for π‘₯, we have that the left-hand limit as π‘₯ approaches two from below is equal to two multiplied by two cubed minus three multiplied by two plus 10. That’s two multiplied by eight minus six plus 10 which is equal to 20.

To find the value of the function at two itself, we use the second part of its definition because this is valid for values of π‘₯ greater than or equal to two. Substituting two into this part of the definition gives π‘Ž multiplied by two cubed plus seven multiplied by two minus 10. That’s eight π‘Ž plus 14 minus 10 which simplifies to eight π‘Ž plus four. As this part of the function’s definition is valid for π‘₯ values greater than or equal to two, the right-hand limit as π‘₯ approaches two from above will be equal to this same expression. It’s also equal to eight π‘Ž plus four.

Now remember, in order for the function to be continuous at π‘₯ equals two and therefore continuous throughout its domain, we require that the left-hand and right-hand limits are equal to one another. So, we need to set 20 and eight π‘Ž plus four, which is our expression for both the right-hand limit and the value of the function itself at two, equal to one another. Doing so, we have the equation eight π‘Ž plus four is equal to 20, which we can solve to find the value of π‘Ž for which the function will be continuous. We subtract four from each side and then divide by eight to find that π‘Ž is equal to two.

So, by considering the left- and right-hand limits of this function as π‘₯ approaches two from either side and an expression for the function at two itself. And then, setting these expressions equal to one another. We’ve found that the function will be continuous when π‘Ž is equal to two.

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