Problem 78
Question
Use a graphing utility to graph the function. Use the graph to determine whether the function has an inverse that is a function (that is, whether the function is one-to-one). $$f(x)=\frac{x^{3}}{2}$$
Step-by-Step Solution
Verified Answer
Yes, the function \(f(x) = \frac{x^{3}}{2}\) is one-to-one and therefore has an inverse that is also a function.
1Step 1: Understanding the function
The function given is \(f(x) = \frac{x^{3}}{2}\). It is a cubic function which means graphically, it will have a unique curve shape.
2Step 2: Graph the function
Use a graphing utility to plot the graph of \(f(x) = \frac{x^{3}}{2}\). The graph will be a curve that starts from the bottom left, crosses the origin (0,0) and heads upwards to the right. The function is increasing on the whole real line.
3Step 3: Apply the horizontal line test
The horizontal line test is used to determine if a function is one-to-one. If a horizontal line intersects the graph of the function at most one time, the function is one-to-one. In the case of \(f(x) = \frac{x^{3}}{2}\), any horizontal line drawn will intersect the graph at only one point. This means the function passes the horizontal line test and is one-to-one.
4Step 4: Determine if the function has an inverse
Since the function \(f(x) = \frac{x^{3}}{2}\) is one-to-one, we can say that it has an inverse that is also a function. This is based on the property that a function is invertible if and only if it is one-to-one.
Other exercises in this chapter
Problem 77
Give the slope and \(y\) -intercept of each line whose equation is given. Assume that \(B \neq 0\) $$A x+B y=C$$
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Give an example of a circle's equation in standard form Describe how to find the center and radius for this circle.
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Express the given function \(h\) as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=\sqrt{5 x^{2}+3}$$
View solution Problem 78
Begin by graphing the square root function, \(f(x)=\sqrt{x} .\) Then use transformations of this graph to graph the given function. $$h(x)=\sqrt{x+1}-1$$
View solution