Problem 40 - Thomas Calculus in SI Units

Question

a. Show that \(h(x)=x^{3} / 4\) and \(k(x)=(4 x)^{1 / 3}\) are inverses of one another. b. Graph \(h\) and \(k\) over an \(x\) -interval large enough to show the graphs intersecting at (2,2) and \((-2,-2) .\) Be sure the picture shows the required symmetry about the line \(y=x\) c. Find the slopes of the tangents to the graphs of \(h\) and \(k\) at (2,2) and (-2,-2) d. What lines are tangent to the curves at the origin?

Step-by-Step Solution

Answer

The functions are inverses, intersect at (2,2) and (-2,-2), and tangent slopes at these points are 3 and 1/3, respectively. At the origin, \( h(x) \) has a horizontal tangent and \( k(x) \) is not differentiable.

Step 1: Verify if \( h(x) \) and \( k(x) \) are inverses

To show that two functions are inverses, verify if \( h(k(x)) = x \) and \( k(h(x)) = x \).First, calculate \( h(k(x)) = h((4x)^{1/3}) \):\[ h(k(x)) = \frac{((4x)^{1/3})^3}{4} = \frac{4x}{4} = x \]Now, calculate \( k(h(x)) = k\left(\frac{x^3}{4}\right) \):\[ k(h(x)) = \left(4\left(\frac{x^3}{4}\right)\right)^{1/3} = x \]Since both compositions return \( x \), the functions are inverses of each other.

Step 2: Plot graphs of \( h(x) \) and \( k(x) \)

Graph \( h(x) = \frac{x^3}{4} \) and \( k(x) = (4x)^{1/3} \) over an interval covering \(-2 \leq x \leq 2\).Observe the symmetry about the line \( y = x \), confirming that the functions intersect at points (2,2) and (-2,-2), illustrating that they are inverses.

Step 3: Find tangent slopes at (2,2)

For \( h(x) = \frac{x^3}{4} \), find \( h'(x) = \frac{3x^2}{4} \). At \( x=2 \), the slope is \( \frac{3(2)^2}{4} = 3 \).For \( k(x) = (4x)^{1/3} \), find the derivative:\[ k'(x) = \frac{1}{3}(4x)^{-2/3} \times 4 = \frac{4}{3(4x)^{2/3}} \]At \( x=2 \), the slope is:\[ k'(2) = \frac{4}{3(8)^{2/3}} = \frac{4}{12^{2/3}} = \frac{1}{3} \]

Step 4: Find tangent slopes at (-2,-2)

For the slope of \( h(x) \) at \( x=-2 \), use the derivative:\[ h'(-2) = \frac{3(-2)^2}{4} = 3 \]For \( k(x) \) at \( x = -2 \), calculate:\[ k'(-2) = \frac{4}{3((-8)^{2/3})} = \frac{1}{3} \]

Step 5: Tangent lines at the origin

At the origin, where \( (0,0) \), the derivative for both functions can show the slope:The slope for \( h(x) \) is \( h'(0) = 0 \).The slope for \( k(x) \) is not defined at \( x=0 \) since \( (4x)^{-2/3} \) becomes undefined.Thus, the tangent line for \( h(x) \) at the origin is horizontal, and no tangent exists for \( k(x) \) due to it not being differentiable at \( x=0 \).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. This line represents the instantaneous rate of change of the curve at that precise location. For example, in this exercise, we found the slopes of the tangent lines for both functions,

• For the function \( h(x) = \frac{x^3}{4} \), the slope of the tangent line at \((2,2)\) and \((-2,-2)\) is 3.
• For \( k(x) = (4x)^{1/3} \), the slope at these points is 1/3.

Notice how the tangent line’s slope gives us a direct insight into how steep the curve is at those points. Additionally, determining tangent lines at critical points, such as the origin, can illuminate special characteristics of the functions. At the origin, the tangent line for \( h(x) \) is horizontal, showcasing a flat slope of zero. But for \( k(x) \), a tangent doesn’t exist, showing the complexity of differentiability at points sometimes.

Function Composition

Function composition involves applying one function to the results of another. In math terms, if you have functions \( f(x) \) and \( g(x) \), the composition is written as \( f(g(x)) \). This method is particularly useful to verify inverse functions.
In the provided problem, to confirm that \( h(x) \) and \( k(x) \) are inverses, both combinations \( h(k(x)) \) and \( k(h(x)) \) were calculated.
To find \( h(k(x)) \):

• Substitute \( k(x) \) into \( h(x) \) results in \( h((4x)^{1/3}) = x \)

For \( k(h(x)) \):

• Substitute \( h(x) \) into \( k(x) \) yields \( k\left(\frac{x^3}{4}\right) = x \)

Both compositions lead back to \( x \), confirming \( h(x) \) and \( k(x) \) are indeed inverses. This property is essential in illustrating the behaviors of complex mathematical relationships.

Derivative

Derivatives represent the rate of change of a function concerning its independent variable. Simply put, it's the function providing the slope of the tangent line to the curve at any point. To compute the derivatives, several rules can be utilized, such as the power rule or chain rule.
Here’s how it is applied:

• For \( h(x) = \frac{x^3}{4} \), via the power rule, the derivative is \( h'(x) = \frac{3x^2}{4} \).
• For \( k(x) = (4x)^{1/3} \), it requires applying the chain rule, resulting in \( k'(x) = \frac{4}{3(4x)^{2/3}} \).

These derivatives help calculate the slopes of tangent lines at specific points.For instance, at \( x=2 \), we know:\

• For \( h(x) \), the slope \( h'(2) = 3 \).
• For \( k(x) \), \( k'(2) = \frac{1}{3} \).

Understanding and using derivatives is critical to exploring the dynamic nature of mathematical functions.

Graph Symmetry

Graph symmetry refers to the balanced and proportional visuals of a function around a central line or point. Symmetry allows us to predict and visualize how a function behaves across different regions of its graph.
In this exercise, symmetry about the line \( y = x \) is crucial because it demonstrates that \( h(x) \) and \( k(x) \) are inverse functions. Their graphs mirror each other over this line. Here's why it's important:

• The functions intersect at points (2,2) and (-2,-2), showcasing their symmetric nature.
• For any point \((a, b)\) on one function, the inverse will pass through \((b, a)\) on the other function, maintaining symmetry about \( y = x \).

Such graphical symmetry aids in confirming the theoretical calculations and understanding the visual implications of inverse relationships. Graph symmetry is an essential tool for comprehending how functions relate and interact visually.