Question

given the function ( f(x) = (x + 11)^3 ), complete parts a through c.
(a) find an equation for ( f^{-1}(x) ).
( f^{-1}(x) = square )
(type an exact answer, using radicals as needed.)
(b) choose the correct graph which shows ( f ) and ( f^{-1} ) graphed in the same coordinate system.
options: a, b, c, d (with graphs)
(c) state the domain and range of ( f ) and ( f^{-1} ) using interval notation.
the domain of ( f(x) ) is ( square ), and the range of ( f(x) ) is ( square ).
the domain of ( f^{-1}(x) ) is ( square ), and the range of ( f^{-1}(x) ) is ( square ).

Explanation:

Step1: Find the inverse function

To find the inverse of \( f(x) = (x + 11)^3 \), we start by letting \( y = (x + 11)^3 \). Then we solve for \( x \) in terms of \( y \). Take the cube root of both sides: \( \sqrt[3]{y} = x + 11 \). Then subtract 11 from both sides: \( x = \sqrt[3]{y} - 11 \). Now we swap \( x \) and \( y \) to get the inverse function: \( f^{-1}(x) = \sqrt[3]{x} - 11 \).

Step2: Analyze the graph of \( f \) and \( f^{-1} \)

The function \( f(x) = (x + 11)^3 \) is a cubic function, which is a one - to - one function (since the derivative \( f^\prime(x)=3(x + 11)^2\geq0\) and is only zero at \( x=-11\), and the function is strictly increasing for \( x
eq - 11\) and increasing at \( x = - 11\) as well). The graph of a function and its inverse are symmetric about the line \( y=x\). The function \( f(x)=(x + 11)^3\) has a horizontal shift of 11 units to the left of the parent function \( y = x^3\). The inverse function \( f^{-1}(x)=\sqrt[3]{x}-11\) has a vertical shift of 11 units down from the parent function \( y=\sqrt[3]{x}\). Looking at the graphs, we need to find the pair of graphs that are symmetric about \( y = x\).

Step3: Find the domain and range of \( f \) and \( f^{-1} \)

For the function \( f(x)=(x + 11)^3\), since it is a cubic function, the domain of a cubic function is all real numbers. In interval notation, the domain of \( f(x)\) is \( (-\infty,\infty) \). The range of a cubic function is also all real numbers, so the range of \( f(x)\) is \( (-\infty,\infty) \).

For the inverse function \( f^{-1}(x)=\sqrt[3]{x}-11\), the cube root function \( y = \sqrt[3]{x}\) has a domain of all real numbers, and when we shift it down by 11 units, the domain of \( f^{-1}(x)\) remains all real numbers, so the domain of \( f^{-1}(x)\) is \( (-\infty,\infty) \). The range of \( f^{-1}(x)\) is also all real numbers, so the range of \( f^{-1}(x)\) is \( (-\infty,\infty) \).

Part (a)

Answer:

\( f^{-1}(x)=\sqrt[3]{x}-11 \)

Part (b)

To determine the correct graph, we know that the graph of \( f(x)=(x + 11)^3\) and \( f^{-1}(x)=\sqrt[3]{x}-11\) should be symmetric about \( y = x\). We can also analyze the key points. For \( f(x)=(x + 11)^3\), when \( x=-11\), \( f(-11) = 0\). For \( f^{-1}(x)=\sqrt[3]{x}-11\), when \( x = 0\), \( f^{-1}(0)=-11\). The graph that shows this symmetry and the correct key points will be the one where the two graphs are mirror images across \( y=x\). (Since we can't see the exact graphs in detail here, but based on the function properties, we can reason about it. If we assume the graphs are labeled as per the standard, we would pick the graph where \( f(x)\) is a cubic shifted left and \( f^{-1}(x)\) is a cube - root shifted down and they are symmetric about \( y = x\))

Part (c)

The domain of \( f(x)\) is \( (-\infty,\infty) \), the range of \( f(x)\) is \( (-\infty,\infty) \). The domain of \( f^{-1}(x)\) is \( (-\infty,\infty) \), the range of \( f^{-1}(x)\) is \( (-\infty,\infty) \)