Problem 39
Question
\begin{equation} \begin{array}{l}{\text { a. Show that } f(x)=x^{3} \text { and } g(x)=\sqrt[1]{x} \text { are inverses of one }} \\ {\text { another. }} \\\ {\text { b. Graph } f \text { and } g \text { over an } x \text { -interval large enough to show the }} \\ {\text { graphs intersecting at }(1,1) \text { and }(-1,-1) \text { . Be sure the picture }} \\ {\text { shows the required symmetry about the line } y=x \text { . }}\\\\{\text { c. Find the slopes of the tangent lines to the graphs of } f \text { and } g \text { at }} \\\ {(1,1) \text { and }(-1,-1) \text { (four tangent lines in all). }} \\ {\text { d. What lines are tangent to the curves at the origin? }}\end{array} \end{equation}
Step-by-Step Solution
Verified Answer
f(x) and g(x) are inverses; tangents at (1,1) and (-1,-1) have slopes 3 (f) and 1 (g); at origin, tangent to f is horizontal, to g is y=x.
1Step 1: Check if f(x) and g(x) are inverses
To prove that two functions \( f(x) \) and \( g(x) \) are inverses, we need to show that \( f(g(x)) = x \) and \( g(f(x)) = x \). Start by evaluating \( f(g(x)) \). Given \( g(x) = x \), then \( f(g(x)) = f(x) = x^3 \). Now evaluate \( g(f(x)) \), which results in \( g(x^3) = x \). Both satisfy the conditions, so \( f(x) \) and \( g(x) \) are indeed inverses.
2Step 2: Graph f(x) and g(x)
Graph \( f(x) = x^3 \) and \( g(x) = x \) over the interval, for example \(-2 \leq x \leq 2\). The graph of \( f(x) \) is a cubic curve and \( g(x) \) is a line with a 45-degree angle to the axes, intersecting the cubic curve at points \((1,1)\) and \((-1,-1)\). The graphs should be symmetrical about the line \( y = x \).
3Step 3: Find tangent slopes at (1,1) and (-1,-1)
The slope of the tangent line to a curve at a point is given by the derivative of the function at that point. For \( f(x) = x^3 \), the derivative is \( f'(x) = 3x^2 \). At \( x = 1 \), the slope is \( 3 \), and at \( x = -1 \), it is also \( 3 \). The slope of the tangent line for \( g(x) = x \) is \( g'(x) = 1 \). At \( (1, 1) \) and \( (-1, -1) \), the slopes are both \( 1 \).
4Step 4: Find tangent lines at the origin
At the origin \((0, 0)\), the slope of the tangent line for \( f(x) = x^3 \) using \( f'(0) = 0 \) is horizontal. For \( g(x) = x \), since its derivative is always \( 1 \), the tangent at the origin is a line of slope \( 1 \), which is \( y = x \).
Key Concepts
Cubic FunctionsDerivativesFunction GraphsTangent Lines
Cubic Functions
Cubic functions are polynomials of degree three and have the form \( f(x) = ax^3 + bx^2 + cx + d \), where \( a eq 0 \). These functions are known for their distinct 'S' shaped curve and can cross the x-axis up to three times. The classic example of a cubic function is \( f(x) = x^3 \), which is simple yet exhibits the key characteristics of cubic behavior.
- They always have at least one real root (since they are polynomials of odd degree).
- They can have one or two turning points, requiring calculus to locate them.
- The graph of a cubic function might have points of inflection, where the curve changes concavity.
Derivatives
Derivatives represent the rate of change of a function with respect to a variable. They are essential for finding slopes of tangent lines to curves at specific points.
For a cubic function like \( f(x) = x^3 \), the derivative \( f'(x) = 3x^2 \) gives us the slope of the tangent line at any x-value.
For a cubic function like \( f(x) = x^3 \), the derivative \( f'(x) = 3x^2 \) gives us the slope of the tangent line at any x-value.
- The derivative is a fundamental tool in calculus, used to compute instantaneous rates of change.
- To find the slope of the tangent line at a particular point, plug the x-value into the derivative expression.
- For example, at \( x = 1 \), \( f'(1) = 3(1)^2 = 3 \), indicating the slope of the tangent line is 3.
Function Graphs
Graphing functions provides a visual representation of mathematical relationships. For the functions \( f(x) = x^3 \) and \( g(x) = x \), graphing helps illuminate their properties and behaviors.
- The cubic function \( f(x) = x^3 \) displays a steep curve increasing exponentially as x moves away from zero in either direction.
- The line \( g(x) = x \) represents a simple linear relationship with an angle of 45 degrees to both axes.
- These functions intersect at points where their values and derivatives may tell us more about their relationships, specifically at \((1, 1)\) and \((-1, -1)\).
Tangent Lines
Tangent lines are straight lines that touch a curve at only one point and have the same slope as the curve did at that point. For any function graph, the tangent line provides a linear approximation to the curve near the point of tangency.
- To find the tangent line at a point, calculate the derivative and evaluate it at that point to find the slope.
- The equation of the tangent line can be written in point-slope form: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \((x_1, y_1)\) is the point of tangency.
- At the origin, the tangent for \( f(x) = x^3 \) is horizontal with a slope of 0, while for \( g(x) = x \), it is \( y = x \), showing how derivatives differ in effect between linear and cubic functions.
Other exercises in this chapter
Problem 39
Evaluate the integrals. \begin{equation}\int_{\ln 4}^{\ln 9} e^{x / 2} d x\end{equation}
View solution Problem 39
Evaluate the integrals in Exercises \(39-56\) $$\int_{-3}^{-2} \frac{d x}{x}$$
View solution Problem 40
In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\cot ^{-1} \frac{1}{x}-\tan ^{-1} x $$
View solution Problem 40
Evaluate the integrals. \begin{equation}\int_{0}^{\ln 16} e^{x / 4} d x\end{equation}
View solution