Video Transcript
Given the function 𝑓 of 𝑥 is equal to three 𝑥 minus one when 𝑥 is greater than or equal to negative one and 𝑥 is less than 10 and 𝑓 of 𝑥 is equal to two 𝑥 plus 11 when 𝑥 is greater than or equal to 10, find the value of 𝑎 such that 𝑓 evaluated at 𝑎 is equal to two.
In this question, we’re given a piecewise-defined function 𝑓 of 𝑥, and we need to determine any value of 𝑎 such that 𝑓 evaluated at 𝑎 is equal to two. To do this, we recall that piecewise-defined functions have different definitions depending on our input value of 𝑥. These are denoted by the two inequalities on the right of our piecewise definition. These are called the subdomains of our function 𝑓 of 𝑥.
First, if our input value of 𝑥 is greater than or equal to negative one and less than 10, then our function 𝑓 of 𝑥 is equal to the function three 𝑥 minus one. Second, if our input value of 𝑥 is greater than or equal to 10, then our function 𝑓 of 𝑥 is equal to two 𝑥 plus 11. Therefore, there are two possible ways we can input a value of 𝑎 such that 𝑓 evaluated at 𝑎 is equal to two. We could input a value of 𝑥 which satisfies the first inequality such that three 𝑥 minus one is equal to two. Or we could input a value of 𝑥 which satisfies the second inequality such that two 𝑥 plus 11 is equal to two. In both cases, we’re going to need to find the input values of 𝑥, which make each subfunction equal to two.
Let’s start with our first subfunction. We want to solve three 𝑥 minus one is equal to two. We add one to both sides of our equation to get three 𝑥 is equal to three, and then we divide our equation through by three to see that 𝑥 is equal to one. So, when 𝑥 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 𝑥 minus one at this input value of 𝑥. 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 𝑥. 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 𝑎.
But remember, we still need to check the second subfunction. We want two 𝑥 plus 11 to be equal to two. We then solve this equation for 𝑥. We subtract 11 from both sides of the equation to get two 𝑥 is equal to negative nine. Then, we divide both sides of the equation through by two. We get 𝑥 is equal to negative nine divided by two. So, if we substitute 𝑥 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 𝑥 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 𝑎. So, the only possible value for the constant 𝑎 is one.
Therefore, we were able to show if 𝑓 of 𝑥 is equal to three 𝑥 minus one when 𝑥 is greater than or equal to negative one and 𝑥 is less than 10 and 𝑓 of 𝑥 is equal to two 𝑥 plus 11 when 𝑥 is greater than or equal to 10, then if we’re told that 𝑓 evaluated at 𝑎 is equal to two, we’ve shown the only possible value of 𝑎 is one.
