Question

given the function f(x)=x³ - 4, 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 \( f(x) = x^3 - 4 \). Replace \( f(x) \) with \( y \), so we have \( y = x^3 - 4 \).

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

To find the inverse function, we swap the roles of \( x \) and \( y \). This gives us \( x = y^3 - 4 \).

Step3: Solve for \( y \)

We need to isolate \( y \) on one side of the equation. First, add 4 to both sides: \( x + 4 = y^3 \). Then, take the cube root of both sides to solve for \( y \). The cube root of \( y^3 \) is \( y \), and the cube root of \( x + 4 \) is \( \sqrt[3]{x + 4} \). So we get \( y = \sqrt[3]{x + 4} \).

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

Now that we have solved for \( y \) in terms of \( x \), we replace \( y \) with \( f^{-1}(x) \) to get the inverse function. So \( f^{-1}(x) = \sqrt[3]{x + 4} \).

Step1: Analyze the function \( f(x)=x^3 - 4 \)

The function \( f(x)=x^3 - 4 \) is a cubic function. The parent function is \( y = x^3 \), which has a point - symmetric graph about the origin. The graph of \( f(x)=x^3 - 4 \) is the graph of \( y = x^3 \) shifted down 4 units. We can find some key points: when \( x = 0 \), \( f(0)=0^3 - 4=-4 \); when \( x = 1 \), \( f(1)=1^3 - 4=-3 \); when \( x=-1 \), \( f(-1)=(-1)^3 - 4=-5 \); when \( x = 2 \), \( f(2)=2^3 - 4 = 4 \); when \( x=-2 \), \( f(-2)=(-2)^3 - 4=-12 \).

Step2: Analyze the inverse function \( f^{-1}(x)=\sqrt[3]{x + 4} \)

The inverse function \( f^{-1}(x)=\sqrt[3]{x + 4} \) can be thought of as a transformation of the parent cube - root function \( y=\sqrt[3]{x} \). The graph of \( y=\sqrt[3]{x + 4} \) is the graph of \( y=\sqrt[3]{x} \) shifted left 4 units. Some key points: when \( x=-4 \), \( f^{-1}(-4)=\sqrt[3]{-4 + 4}=0 \); when \( x=-3 \), \( f^{-1}(-3)=\sqrt[3]{-3 + 4}=1 \); when \( x=-5 \), \( f^{-1}(-5)=\sqrt[3]{-5 + 4}=-1 \); when \( x = 4 \), \( f^{-1}(4)=\sqrt[3]{4 + 4}=\sqrt[3]{8}=2 \); when \( x=-12 \), \( f^{-1}(-12)=\sqrt[3]{-12 + 4}=\sqrt[3]{-8}=-2 \).

Step3: Graph the functions

We know that the graph of a function and its inverse are symmetric about the line \( y = x \). So we can plot the points for \( f(x) \) and then plot the points for \( f^{-1}(x) \) such that they are symmetric about \( y = x \). For example, the point \( (0,-4) \) on \( f(x) \) corresponds to the point \( (-4,0) \) on \( f^{-1}(x) \), the point \( (1,-3) \) on \( f(x) \) corresponds to the point \( (-3,1) \) on \( f^{-1}(x) \), and so on. Then we can draw the smooth curves for both functions.

Step1: Domain and range of \( f(x)=x^3 - 4 \)

For the function \( f(x)=x^3 - 4 \), since we can cube any real number \( x \) (there are no restrictions on the input \( x \) for the cubic function), the domain of \( f(x) \) is all real numbers. In interval notation, the domain of \( f(x) \) is \( (-\infty,\infty) \).
For the range of \( f(x)=x^3 - 4 \), as \( x \) takes on all real values, \( x^3 \) also takes on all real values (because the cubic function \( y = x^3 \) is one - to - one and onto the set of real numbers). When we subtract 4 from \( x^3 \), the range of \( f(x)=x^3 - 4 \) is also all real numbers. In interval notation, the range of \( f(x) \) is \( (-\infty,\infty) \).

Step2: Domain and range of \( f^{-1}(x)=\sqrt[3]{x + 4} \)

For the inverse function \( f^{-1}(x)=\sqrt[3]{x + 4} \), we can take the cube root of any real number \( x+4 \) (because the cube root function is defined for all real numbers). So the domain of \( f^{-1}(x) \) is all real numbers. In interval notation, the domain of \( f^{-1}(x) \) is \( (-\infty,\infty) \).
The range of an inverse function is equal to the domain of the original function. Since the domain of \( f(x) \) is \( (-\infty,\infty) \), the range of \( f^{-1}(x) \) is also \( (-\infty,\infty) \).

Answer:

(for part a):
\( f^{-1}(x)=\sqrt[3]{x + 4} \)

Part (b)