+2446 #10, why not?
According to the original relation, \(f(x)=\sqrt[3]{x-4}\). There are a few steps that need to be done in order to find the inverse.
#1) Convert to y=-notation. I find that it is easier when working with this notation. This is quite a simple step, wouldn't you agree?
\(f(x)=\sqrt[3]{x-4}\Rightarrow y=\sqrt[3]{x-4}\)
#2) Now, replace all instances of a "y" with an "x," and replace all instances of an "x" with a "y." This is not a difficult step, either, as you might imagine.
\(y=\sqrt[3]{x-4}\Rightarrow x=\sqrt[3]{y-4}\)
#3) Solve for y. This step can range in difficulty. In this case, it is quite a simple step.
| \(x=\sqrt[3]{y-4}\) | Cube both sides. |
| \(x^3=y-4\) | Add 4 to both sides to isolate y. |
| \(y=x^3+4\) | |
#4) Convert back to function notation since the original problem was given in function notation. This is quite simple, too.
\(y=x^3+4\Rightarrow f^{-1}(x)=x^3+4\)
#5) Correspond your answer with the answer choices given!
+2446 #10, why not?
According to the original relation, \(f(x)=\sqrt[3]{x-4}\). There are a few steps that need to be done in order to find the inverse.
#1) Convert to y=-notation. I find that it is easier when working with this notation. This is quite a simple step, wouldn't you agree?
\(f(x)=\sqrt[3]{x-4}\Rightarrow y=\sqrt[3]{x-4}\)
#2) Now, replace all instances of a "y" with an "x," and replace all instances of an "x" with a "y." This is not a difficult step, either, as you might imagine.
\(y=\sqrt[3]{x-4}\Rightarrow x=\sqrt[3]{y-4}\)
#3) Solve for y. This step can range in difficulty. In this case, it is quite a simple step.
| \(x=\sqrt[3]{y-4}\) | Cube both sides. |
| \(x^3=y-4\) | Add 4 to both sides to isolate y. |
| \(y=x^3+4\) | |
#4) Convert back to function notation since the original problem was given in function notation. This is quite simple, too.
\(y=x^3+4\Rightarrow f^{-1}(x)=x^3+4\)
#5) Correspond your answer with the answer choices given!
Help.
