Problem 29 - Thomas Calculus

Question

a. Show that \(f(x)=x^{3}\) and \(g(x)=\sqrt[3]{x}\) are inverses of one another. b. Graph \(f\) and \(g\) over an \(x\) -interval large enough to show the graphs intersecting at \((1,1)\) and \((-1,-1) .\) Be sure the picture shows the required symmetry about the line \(y=x\) . c. Find the slopes of the tangents to the graphs of \(f\) and \(g\) at \((1,1)\) and \((-1,-1)\) (four tangents in all). d. What lines are tangent to the curves at the origin?

Step-by-Step Solution

Answer

Functions \( f(x) = x^3 \) and \( g(x) = \sqrt[3]{x} \) are inverses; they intersect at \((1,1)\) and \((-1,-1)\) with slopes 3 and \( \frac{1}{3} \); tangent at origin is \( y = 0 \).

Step 1: Inverse Check with Composition

To show that functions are inverses, compose them and see if they return the identity function. First, calculate \( (f \circ g)(x) \) which means \( f(g(x)) \). Since \( g(x) = \sqrt[3]{x} \), we have \( f(g(x)) = (\sqrt[3]{x})^3 = x \), which simplifies to \( x \). This confirms that \( f(g(x)) = x \). Next, calculate \( (g \circ f)(x) \) which means \( g(f(x)) \). Since \( f(x) = x^3 \), then \( g(f(x)) = \sqrt[3]{x^3} = x \). This confirms that \( g(f(x)) = x \). Since both compositions return \( x \), \( f \) and \( g \) are inverses.

Step 2: Graphing the Functions

To graph \( f \) and \( g \), plot \( y = x^3 \) and \( y = \sqrt[3]{x} \) over a suitable range, such as \([-5, 5]\). The graphs will intersect at points \((1, 1)\) and \((-1, -1)\). Both graphs exhibit symmetry about the line \( y = x \). Ensure the graph includes this symmetry by checking that any point \((a, b)\) also includes the point \((b, a)\).

Step 3: Find Slopes at (1,1) and (-1,-1)

First, find the derivative of \( f(x) = x^3 \), which is \( f'(x) = 3x^2 \). Evaluate \( f'(1) = 3 \cdot 1^2 = 3 \) and \( f'(-1) = 3 \cdot (-1)^2 = 3 \). Now, find the derivative of \( g(x) = \sqrt[3]{x} = x^{1/3} \), which is \( g'(x) = \frac{1}{3}x^{-2/3} \). Evaluate \( g'(1) = \frac{1}{3} \cdot 1^{-2/3} = \frac{1}{3} \) and \( g'(-1) = \frac{1}{3} \cdot (-1)^{-2/3} = \frac{1}{3} \). Thus, the slopes of the tangents at \((1,1)\) and \((-1,-1)\) for \( f \) and \( g \) are 3 and \( \frac{1}{3} \) respectively.

Step 4: Tangents at the Origin

The tangent to \( f(x) = x^3 \) at \((0,0)\) is found by evaluating its derivative at \( x = 0\). \( f'(x) = 3x^2 \) gives \( f'(0) = 3 \cdot 0^2 = 0 \), so the tangent line is horizontal, i.e., \( y = 0 \). For \( g(x) = \sqrt[3]{x} \), \( g'(x) = \frac{1}{3}x^{-2/3} \) gives \( g'(0) \) as undefined due to division by zero; however, the curve is symmetric and also has a horizontal tangent \( y = 0 \) at the origin by the symmetry argument.

Key Concepts

Composition of Functions

When dealing with inverse functions, the composition of functions plays a crucial role. The basic idea is to compose two functions, say, \( f(x) \) and \( g(x) \), such that the output of one function becomes the input to the other.
For functions to be true inverses, composing them should result in the identity function, \( x \). Let’s break this down:

• Calculate \( (f \circ g)(x) = f(g(x)) \). For \( f(x) = x^3 \) and \( g(x) = \sqrt[3]{x} \), this composition simplifies to \( ((\sqrt[3]{x})^3) = x \).
• Similarly, calculate \( (g \circ f)(x) = g(f(x)) \). This gives \( \sqrt[3]{x^3} = x \).
• If both compositions equal \( x \), then \( f(x) \) and \( g(x) \) are confirmed to be inverses of each other.

Understanding and verifying this concept through the composition is fundamental because it shows how two functions "undo" each other's operations.

Graphing Functions

Graphing functions helps in visually interpreting mathematical relationships. When graphing inverse functions, it is critical to see how the graphs relate to each other. For \( f(x) = x^3 \) and \( g(x) = \sqrt[3]{x} \) over a range like \([-5, 5]\), there are a few key observations:

• The graphs of \( f \) and \( g \) should intersect at points where the functions have the same value, such as \((1, 1)\) and \((-1, -1)\).
• Both graphs should exhibit symmetry around the line \( y = x \). This means that flipping the coordinates \((a, b)\) will give \((b, a)\).
• Symmetry around \( y = x \) is a hallmark of inverse functions, and this graphical representation confirms their inverse nature.

This visual approach can help students grasp how functions and their inverses relate in a geometric context.

Derivative of a Function

Finding derivatives is essential for understanding the behavior of functions, especially when it comes to rates of change and determining tangent lines. Here's how we approached derivatives for our functions:

• The derivative of \( f(x) = x^3 \) is \( f'(x) = 3x^2 \). It tells us the slope of the tangent line at any point \( x \).
• For instance, \( f'(1) = 3 \) and \( f'(-1) = 3 \), indicating consistent slopes at points \( (1,1) \) and \( (-1,-1) \).
• The derivative of \( g(x) = \sqrt[3]{x} \) is \( g'(x) = \frac{1}{3}x^{-2/3} \). At the points \( (1,1) \) and \( (-1,-1)\), \( g'(1) = \frac{1}{3} \) and \( g'(-1) = \frac{1}{3} \), offering us the slopes of tangents there.

By calculating derivatives, we obtain valuable insights into how rapidly or slowly a function changes at given points.

Tangent Lines

Tangent lines are straight lines that touch curves at one point without crossing over. They provide a snapshot of a curve’s slope at a specific point. Let's look at how tangent lines work for our functions:
The slopes of these tangents, calculated using derivatives, give us valuable information:

• For \( f(x) = x^3 \), the tangent line at \((0,0)\) has a slope of 0, thus the line is horizontal, which is \( y = 0 \).
• The function \( g(x) = \sqrt[3]{x} \) presents an undefined tangent slope at \( (0,0) \) but due to symmetry, we know the tangent at this point is also \( y = 0 \).
• Understanding tangent lines in this context is crucial since it shows how these functions and their inverses behave at specific intersections.

Tangent lines paint a clear picture of function behavior, highlighting crucial intersections and providing clearer understanding through derivative analysis.