Question Video: Finding the Value of a Constant in the Domain of a Piecewise-Defined Function

Given the function 𝑓(π‘₯) = 3π‘₯ βˆ’ 1, βˆ’1 ≀ π‘₯ < 10 and 𝑓(π‘₯) = 2π‘₯ + 11, π‘₯ β‰₯ 10, find the value of π‘Ž such that 𝑓(π‘Ž) = 2.

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Video Transcript

Given the function 𝑓 of π‘₯ is equal to three π‘₯ minus one when π‘₯ is greater than or equal to negative one and π‘₯ is less than 10 and 𝑓 of π‘₯ is equal to two π‘₯ plus 11 when π‘₯ is greater than or equal to 10, find the value of π‘Ž such that 𝑓 evaluated at π‘Ž is equal to two.

In this question, we’re given a piecewise-defined function 𝑓 of π‘₯, and we need to determine any value of π‘Ž such that 𝑓 evaluated at π‘Ž is equal to two. To do this, we recall that piecewise-defined functions have different definitions depending on our input value of π‘₯. These are denoted by the two inequalities on the right of our piecewise definition. These are called the subdomains of our function 𝑓 of π‘₯.

First, if our input value of π‘₯ is greater than or equal to negative one and less than 10, then our function 𝑓 of π‘₯ is equal to the function three π‘₯ minus one. Second, if our input value of π‘₯ is greater than or equal to 10, then our function 𝑓 of π‘₯ is equal to two π‘₯ plus 11. Therefore, there are two possible ways we can input a value of π‘Ž such that 𝑓 evaluated at π‘Ž is equal to two. We could input a value of π‘₯ which satisfies the first inequality such that three π‘₯ minus one is equal to two. Or we could input a value of π‘₯ which satisfies the second inequality such that two π‘₯ plus 11 is equal to two. In both cases, we’re going to need to find the input values of π‘₯, which make each subfunction equal to two.

Let’s start with our first subfunction. We want to solve three π‘₯ minus one is equal to two. We add one to both sides of our equation to get three π‘₯ is equal to three, and then we divide our equation through by three to see that π‘₯ is equal to one. So, when π‘₯ is equal to one, our first subfunction will output a value of two. But remember, we need to check that this value is in the first subdomain. Otherwise, our function would not be equal to three π‘₯ minus one at this input value of π‘₯. However, we can see that one is greater than or equal to negative one and one is less than 10, so one is in the first subdomain of 𝑓 of π‘₯. Therefore, 𝑓 evaluated at one is equal to three times one minus one, which we can see is equal to two. So, one is a possible value of π‘Ž.

But remember, we still need to check the second subfunction. We want two π‘₯ plus 11 to be equal to two. We then solve this equation for π‘₯. We subtract 11 from both sides of the equation to get two π‘₯ is equal to negative nine. Then, we divide both sides of the equation through by two. We get π‘₯ is equal to negative nine divided by two. So, if we substitute π‘₯ is equal to nine over two into this subfunction, we will get an output value of two. However, we still need to check that this input value of π‘₯ is in the second subdomain. But negative nine over two is less than 10; it’s not greater than or equal to 10. So, negative nine over two is not in this subdomain, which means it’s not a valid value of π‘Ž. So, the only possible value for the constant π‘Ž is one.

Therefore, we were able to show if 𝑓 of π‘₯ is equal to three π‘₯ minus one when π‘₯ is greater than or equal to negative one and π‘₯ is less than 10 and 𝑓 of π‘₯ is equal to two π‘₯ plus 11 when π‘₯ is greater than or equal to 10, then if we’re told that 𝑓 evaluated at π‘Ž is equal to two, we’ve shown the only possible value of π‘Ž is one.

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