Problem 61
Question
Find the inverse of each one-to-one function. Then graph the function and its inverse in a square window. $$ f(x)=\sqrt[3]{x+3} $$
Step-by-Step Solution
Verified Answer
The inverse is \( f^{-1}(x) = x^3 - 3 \).
1Step 1: Set Function as Equation
To find the inverse of the function, start by writing it as an equation with "y" in place of the function, i.e., \( y = \sqrt[3]{x + 3} \).
2Step 2: Swap Variables
As the definition of inverse function involves switching the roles of x and y, swap the x and y in the equation to get \( x = \sqrt[3]{y + 3} \).
3Step 3: Solve for New Function
Solve for \( y \) in terms of \( x \) from \( x = \sqrt[3]{y + 3} \). Start by cubing both sides to eliminate the cube root, which results in \( x^3 = y + 3 \).
4Step 4: Isolate y
Subtract 3 from both sides to isolate \( y \), yielding \( y = x^3 - 3 \).
5Step 5: Write Inverse Function
Replace \( y \) with \( f^{-1}(x) \) to express the inverse function as \( f^{-1}(x) = x^3 - 3 \).
6Step 6: Graph the Function and its Inverse
To graph \( f(x) = \sqrt[3]{x+3} \) and \( f^{-1}(x) = x^3 - 3 \), note that both graphs should reflect about the line \( y = x \). Use a graphing tool or plot by calculating points. The original function will curve up and the inverse will curve down.
Key Concepts
One-to-One FunctionsGraphing FunctionsCubic Functions
One-to-One Functions
A one-to-one function is a special type of function where each output value corresponds to exactly one input value. This unique characteristic makes it possible for a function to have an inverse.
When learning about inverse functions, identifying whether a function is one-to-one is a crucial first step.
When learning about inverse functions, identifying whether a function is one-to-one is a crucial first step.
- For a function to be one-to-one, each horizontal line should intersect the graph of the function at most once. This is known as the Horizontal Line Test.
- If a function passes this test, it means that no two input values produce the same output, and the function has an inverse.
- The inverse function essentially "reverses" the operations of the original function, swapping roles of inputs and outputs.
Graphing Functions
Graphing functions provides a powerful visual representation of mathematical relationships. In this context, understanding the graphs of both a function and its inverse helps reinforce their connection.
- The graph of the original function \( f(x) = \sqrt[3]{x + 3} \) depicts a standard cubic root curve, shifted horizontally by 3 units due to the \( x+3 \) inside the cube root.
- The graph of the inverse \( f^{-1}(x) = x^3 - 3 \) is a cubic function shifted down by 3 units.
- One vital property of inverse functions is that their graphs are reflections of each other across the line \( y = x \). This means each point \((a, b)\) on the original function corresponds to a point \((b, a)\) on its inverse.
Cubic Functions
Cubic functions are polynomial functions of degree three, typically expressed in the form \( f(x) = ax^3 + bx^2 + cx + d \). They are fundamental in mathematics due to their distinct shape and properties.
- These functions have a characteristic "S" or "N" shaped curve. Depending on the coefficient \(a\), the curve either rises or falls steeply, before leveling out, and then rising or falling steeply again.
- Cubic functions can have one real root or three real roots, and their graphs can feature inflection points where the curve changes concavity.
- In our exercise, the function \( f^{-1}(x) = x^3 - 3 \) is such a cubic function. It presents a single inflection point, reflects symmetry typical of cubic graphs, and displays internal axes shifts due to the \(-3\).
Other exercises in this chapter
Problem 61
Solve. $$ \log _{x} 100=2 $$
View solution Problem 61
Solve for \(x\). $$ 6 x-3(2-5 x)=6 $$
View solution Problem 61
Explain why the graph of an exponential function \(y=b^{x}\) contains the point \((1, b)\).
View solution Problem 61
If \(\log _{b} 2=0.43\) and \(\log _{b} 3=0.68\), evaluate each expression. $$ \log _{b} \frac{3}{9} $$
View solution