# Question Video: Finding the Value of a Constant in the Domain of a Piecewise-Defined Function Mathematics • 10th Grade

Given the function π(π₯) = 3π₯ β 1, β1 β€ π₯ < 10 and π(π₯) = 2π₯ + 11, π₯ β₯ 10, find the value of π such that π(π) = 2.

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### Video Transcript

Given the function π 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 π of π₯ is equal to two π₯ plus 11 when π₯ is greater than or equal to 10, find the value of π such that π evaluated at π is equal to two.

In this question, weβre given a piecewise-defined function π of π₯, and we need to determine any value of π such that π evaluated at π 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 π of π₯.

First, if our input value of π₯ is greater than or equal to negative one and less than 10, then our function π 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 π of π₯ is equal to two π₯ plus 11. Therefore, there are two possible ways we can input a value of π such that π evaluated at π 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, weβre going to need to find the input values of π₯, which make each subfunction equal to two.

Letβ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 π of π₯. Therefore, π 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.

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; itβs not greater than or equal to 10. So, negative nine over two is not in this subdomain, which means itβs not a valid value of π. So, the only possible value for the constant π is one.

Therefore, we were able to show if π 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 π of π₯ is equal to two π₯ plus 11 when π₯ is greater than or equal to 10, then if weβre told that π evaluated at π is equal to two, weβve shown the only possible value of π is one.