Question

given the function f(x) = (x + 1)^3, complete parts a through c. (a) find an equation for f^{-1}(x). (b) graph f and f^{-1} in the same rectangular coordinate system. (c) use interval notation to give the domain and the range of f and f^{-1}.

Explanation:

Part (a)

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x)=(x + 1)^3 \). Replace \( f(x) \) with \( y \), so we have \( y=(x + 1)^3 \).

Step2: Swap \( x \) and \( y \)

To find the inverse function, we swap the roles of \( x \) and \( y \). This gives us \( x=(y + 1)^3 \).

Step3: Solve for \( y \)

Take the cube - root of both sides of the equation \( x=(y + 1)^3 \). The cube - root of \( x \) is \( \sqrt[3]{x} \), and the cube - root of \( (y + 1)^3 \) is \( y + 1 \). So we get \( \sqrt[3]{x}=y + 1 \). Then, subtract 1 from both sides to solve for \( y \): \( y=\sqrt[3]{x}-1 \).

Step4: Replace \( y \) with \( f^{-1}(x) \)

We replace \( y \) with \( f^{-1}(x) \), so the inverse function \( f^{-1}(x)=\sqrt[3]{x}-1 \).

Part (b)

Step1: Analyze the graph of \( f(x)=(x + 1)^3 \)

The function \( y = f(x)=(x + 1)^3 \) is a cubic function. The parent function of a cubic function is \( y=x^3 \). The graph of \( y=(x + 1)^3 \) is the graph of \( y = x^3 \) shifted 1 unit to the left. The domain and range of \( y=(x + 1)^3 \) are both \( (-\infty,\infty) \) since for any real number \( x \), \( (x + 1)^3 \) is defined, and we can get any real number as an output by choosing an appropriate \( x \).

Step2: Analyze the graph of \( f^{-1}(x)=\sqrt[3]{x}-1 \)

The function \( y = f^{-1}(x)=\sqrt[3]{x}-1 \) is also a cubic - type function (it is the inverse of a cubic function). The parent function is \( y=\sqrt[3]{x} \). The graph of \( y=\sqrt[3]{x}-1 \) is the graph of \( y=\sqrt[3]{x} \) shifted 1 unit down. The domain and range of \( y=\sqrt[3]{x}-1 \) are both \( (-\infty,\infty) \) because we can take the cube - root of any real number \( x \), and we can get any real number as an output.

Step3: Graphing in the same coordinate system

• For \( f(x)=(x + 1)^3 \):
• When \( x=-2 \), \( f(-2)=(-2 + 1)^3=(-1)^3=-1 \)
• When \( x=-1 \), \( f(-1)=(-1 + 1)^3=0^3 = 0 \)
• When \( x = 0 \), \( f(0)=(0 + 1)^3=1 \)
• For \( f^{-1}(x)=\sqrt[3]{x}-1 \):
• When \( x=-1 \), \( f^{-1}(-1)=\sqrt[3]{-1}-1=-1 - 1=-2 \)
• When \( x = 0 \), \( f^{-1}(0)=\sqrt[3]{0}-1=0 - 1=-1 \)
• When \( x = 1 \), \( f^{-1}(1)=\sqrt[3]{1}-1=1 - 1=0 \)

We can plot these points and draw the curves. Also, we know that the graph of a function and its inverse are symmetric about the line \( y = x \). So we can use this symmetry to help us graph \( f^{-1}(x) \) once we have the graph of \( f(x) \).

Part (c)

Step1: Domain and range of \( f(x)=(x + 1)^3 \)

For the function \( f(x)=(x + 1)^3 \), since we can substitute any real number for \( x \) (there are no restrictions like division by zero or square - root of a negative number), the domain of \( f(x) \) is all real numbers. In interval notation, the domain of \( f(x) \) is \( (-\infty,\infty) \).

To find the range, we note that for any real number \( y \), we can solve the equation \( y=(x + 1)^3 \) for \( x \) (as we did when finding the inverse function). For any real \( y \), \( x=\sqrt[3]{y}-1 \) is a real number. So the range of \( f(x) \) is also all real numbers, and in interval notation, the range of \( f(x) \) is \( (-\infty,\infty) \).

Step2: Domain and range of \( f^{-1}(x)=\sqrt[3]{x}-1 \)

For the inverse function \( f^{-1}(x)=\sqrt[3]{x}-1 \), we can take the cube - root of any real number \( x \) (the cube - root function is defined for all real numbers). So the domain of \( f^{-1}(x) \) is \( (-\infty,\infty) \) in interval notation.

The range of an inverse function is the domain of the original function. Since the domain of \( f(x) \) is \( (-\infty,\infty) \), the range of \( f^{-1}(x) \) is \( (-\infty,\infty) \) in interval notation.

Part (a) Answer:

\( f^{-1}(x)=\sqrt[3]{x}-1 \)

Part (b) (Brief description of graphing)

The graph of \( f(x)=(x + 1)^3 \) is a cubic curve shifted 1 unit left from \( y = x^3 \), and the graph of \( f^{-1}(x)=\sqrt[3]{x}-1 \) is a cubic - root - based curve shifted 1 unit down from \( y=\sqrt[3]{x} \). They are symmetric about the line \( y = x \).

Part (c)

• Domain of \( f \): \( (-\infty,\infty) \)
• Range of \( f \): \( (-\infty,\infty) \)
• Domain of \( f^{-1} \): \( (-\infty,\infty) \)
• Range of \( f^{-1} \): \( (-\infty,\infty) \)

Answer:

Step1: Domain and range of \( f(x)=(x + 1)^3 \)

For the function \( f(x)=(x + 1)^3 \), since we can substitute any real number for \( x \) (there are no restrictions like division by zero or square - root of a negative number), the domain of \( f(x) \) is all real numbers. In interval notation, the domain of \( f(x) \) is \( (-\infty,\infty) \).

To find the range, we note that for any real number \( y \), we can solve the equation \( y=(x + 1)^3 \) for \( x \) (as we did when finding the inverse function). For any real \( y \), \( x=\sqrt[3]{y}-1 \) is a real number. So the range of \( f(x) \) is also all real numbers, and in interval notation, the range of \( f(x) \) is \( (-\infty,\infty) \).

Step2: Domain and range of \( f^{-1}(x)=\sqrt[3]{x}-1 \)

For the inverse function \( f^{-1}(x)=\sqrt[3]{x}-1 \), we can take the cube - root of any real number \( x \) (the cube - root function is defined for all real numbers). So the domain of \( f^{-1}(x) \) is \( (-\infty,\infty) \) in interval notation.

The range of an inverse function is the domain of the original function. Since the domain of \( f(x) \) is \( (-\infty,\infty) \), the range of \( f^{-1}(x) \) is \( (-\infty,\infty) \) in interval notation.

Part (a) Answer:

\( f^{-1}(x)=\sqrt[3]{x}-1 \)

Part (b) (Brief description of graphing)

The graph of \( f(x)=(x + 1)^3 \) is a cubic curve shifted 1 unit left from \( y = x^3 \), and the graph of \( f^{-1}(x)=\sqrt[3]{x}-1 \) is a cubic - root - based curve shifted 1 unit down from \( y=\sqrt[3]{x} \). They are symmetric about the line \( y = x \).

Part (c)

• Domain of \( f \): \( (-\infty,\infty) \)
• Range of \( f \): \( (-\infty,\infty) \)
• Domain of \( f^{-1} \): \( (-\infty,\infty) \)
• Range of \( f^{-1} \): \( (-\infty,\infty) \)