Question

2-134. each of the curves graphed below is not a function. however, restricting the domain or range can turn a non - function into a function. sketch a copy of each graph below onto your paper. then highlight the largest portion of each curve that represents a function. in each case, did you restrict the domain or did you restrict the range? a. b.

Explanation:

Step1: Recall function definition

A function has one - output for each input. For a graph, it must pass the vertical line test.

Step2: Analyze graph a (circle)

For the circle centered at the origin with radius \(r = 4\) (equation \(x^{2}+y^{2}=16\)), we can restrict the domain. If we consider either the upper - half or the lower - half of the circle, it will be a function. The largest portion is either the upper - half \(y=\sqrt{16 - x^{2}}\) or the lower - half \(y =-\sqrt{16 - x^{2}}\). We restrict the domain to \(- 4\leqslant x\leqslant4\) and choose either \(y\geqslant0\) (upper - half) or \(y\leqslant0\) (lower - half).

Step3: Analyze graph b (s - shaped curve)

For the s - shaped curve, we can restrict the domain. We can take a portion of the curve such that for each \(x\) value there is only one \(y\) value. For example, we can take the part of the curve where the curve is either increasing or decreasing monotonically. We restrict the domain to a sub - interval of the \(x\) - values of the curve.

Answer:

a. We restricted the domain. The largest portion can be either the upper - half of the circle \(y=\sqrt{16 - x^{2}}\) with domain \(-4\leqslant x\leqslant4\) or the lower - half \(y =-\sqrt{16 - x^{2}}\) with domain \(-4\leqslant x\leqslant4\).
b. We restricted the domain. We can choose a monotonic part of the s - shaped curve (either increasing or decreasing part) by restricting the \(x\) - values.