Question Video: Determining the Range of a Piecewise-Defined Function given Its Graph Mathematics • 10th Grade

Find the range of the function 𝑓(π‘₯) = π‘₯ + 5 when π‘₯ ∈ [βˆ’5, βˆ’1] and 𝑓(π‘₯) = βˆ’π‘₯ + 3 if π‘₯ ∈ (βˆ’1, 3].

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Video Transcript

Find the range of the function 𝑓 of π‘₯ is equal to π‘₯ plus five when π‘₯ is in the closed interval from negative five to negative one and 𝑓 of π‘₯ is equal to negative π‘₯ plus three if π‘₯ is in the left-open, right-closed interval from negative one to three.

In this question, we’re asked to find the range of a given piecewise-defined function. And we can start by recalling the range of a function is the set of all outputs of that function given the domain of that function, or the set of inputs. And we recall when we graph a function, the 𝑦-coordinate of any point on the curve represents the output value of that function for that given input value of π‘₯. For example, because our function passes through the point negative one, four, we know that 𝑓 evaluated at negative one is equal to four. Four is a possible output of our function. We can even confirm this from our piecewise definition of 𝑓 of π‘₯.

We can see that negative one lies in the first subdomain of our function. So when we evaluate 𝑓 at negative one, we get negative one plus five, which is equal to four. We want to find all of the possible output values of this function. We can see from the graph the highest possible output value is four. Similarly, we can find the lowest possible output value of our function. The lowest possible output value of our function is zero, where it’s worth pointing out because our graph does not have hollow dots at the endpoints on either side, this confirms our function is defined when π‘₯ is equal to negative five and when π‘₯ is equal to three. We can also see this from the subdomains of our function.

So the highest output of our function is four, and the lowest output of our function is zero. And we can see that there are values of π‘₯ for every input in between. And the set of all values from zero to four, including the endpoints, is the closed interval from zero to four. Therefore, we were able to show the range of the function 𝑓 of π‘₯ is equal to π‘₯ plus five when π‘₯ is in the closed interval from negative five to negative one and 𝑓 of π‘₯ is equal to negative π‘₯ plus three when π‘₯ is in the left-open, right-closed interval from negative one to three is the closed interval from zero to four.

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