
How do you find \[{{f}^{-1}}\left( x \right)\] given \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\]
Answer
542.1k+ views
Hint: This type of problem is based on the concept of finding inverse for a function. First, we have to assume the given function as y, that is, \[f\left( x \right)=y\]. Then, make necessary calculations and find the value of x which will be in terms of y by taking the cube root on both the sides of the equation. And then, we have to substitute x in terms of y. Thus, the obtained expression is the required solution, that is, the value of \[{{f}^{-1}}\left( x \right)\] when \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\].
Complete answer:
According to the question, we are asked to find \[{{f}^{-1}}\left( x \right)\] of the given function \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\].
We have been given the function \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\]. -----(1)
We first have to consider \[f\left( x \right)=y\].
We get, \[y=\dfrac{1}{{{x}^{3}}}\].
Using the method of cross-multiplying, that is, \[a=\dfrac{1}{b}\Rightarrow b=\dfrac{1}{a}\].
We get, \[{{x}^{3}}=\dfrac{1}{y}\]. --------(2)
Let us now take the cube root on both the sides of the equation (2).
\[\Rightarrow \sqrt[3]{{{x}^{3}}}=\sqrt[3]{\dfrac{1}{y}}\]
\[\Rightarrow \sqrt[3]{{{x}^{3}}}=\dfrac{\sqrt[3]{1}}{\sqrt[3]{y}}\]
We know that \[\sqrt[3]{{{x}^{3}}}=x\] .
And any power raised to 1 is 1.
We get,
\[\Rightarrow x=\dfrac{1}{\sqrt[3]{y}}\]
But we also know that \[\sqrt[3]{y}={{y}^{\dfrac{1}{3}}}\].
Therefore, \[x=\dfrac{1}{{{y}^{\dfrac{1}{3}}}}\].
We have now obtained the value of x in terms of y.
Now, to find \[{{f}^{-1}}\left( x \right)\] we have to replace y with x.
\[{{f}^{-1}}\left( x \right)\] is nothing but value of x in terms of x in the given function \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\].
\[{{f}^{-1}}\left( x \right)=\dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}}\]
Hence, the value of \[{{f}^{-1}}\left( x \right)\] for the function \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\] is \[\dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}}\].
Note: Whenever you get this type of problem, we should always try to make the necessary calculations in the given equation to get the final of x in terms of x which will be the required answer. We should avoid calculation mistakes based on sign conventions. The final solution can also be written as \[{{f}^{-1}}\left( x \right)=\dfrac{1}{\sqrt[3]{x}}\].
We can check the final answer by this method: \[f\left( {{f}^{-1}}\left( x \right) \right)=x\]
Here \[{{f}^{-1}}\left( x \right)=\dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}}\].
Therefore, \[f\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)=\dfrac{1}{{{\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)}^{3}}}\] [since \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\]]
\[\Rightarrow f\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)=\dfrac{1}{\left( \dfrac{1}{{{x}^{\dfrac{1}{3}\times 3}}} \right)}\]
\[\Rightarrow f\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)=\dfrac{1}{\left( \dfrac{1}{x} \right)}\]
\[\Rightarrow f\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)=x\]
\[\therefore f\left( {{f}^{-1}}\left( x \right) \right)=x\]
Hence, the obtained answer is verified.
Complete answer:
According to the question, we are asked to find \[{{f}^{-1}}\left( x \right)\] of the given function \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\].
We have been given the function \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\]. -----(1)
We first have to consider \[f\left( x \right)=y\].
We get, \[y=\dfrac{1}{{{x}^{3}}}\].
Using the method of cross-multiplying, that is, \[a=\dfrac{1}{b}\Rightarrow b=\dfrac{1}{a}\].
We get, \[{{x}^{3}}=\dfrac{1}{y}\]. --------(2)
Let us now take the cube root on both the sides of the equation (2).
\[\Rightarrow \sqrt[3]{{{x}^{3}}}=\sqrt[3]{\dfrac{1}{y}}\]
\[\Rightarrow \sqrt[3]{{{x}^{3}}}=\dfrac{\sqrt[3]{1}}{\sqrt[3]{y}}\]
We know that \[\sqrt[3]{{{x}^{3}}}=x\] .
And any power raised to 1 is 1.
We get,
\[\Rightarrow x=\dfrac{1}{\sqrt[3]{y}}\]
But we also know that \[\sqrt[3]{y}={{y}^{\dfrac{1}{3}}}\].
Therefore, \[x=\dfrac{1}{{{y}^{\dfrac{1}{3}}}}\].
We have now obtained the value of x in terms of y.
Now, to find \[{{f}^{-1}}\left( x \right)\] we have to replace y with x.
\[{{f}^{-1}}\left( x \right)\] is nothing but value of x in terms of x in the given function \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\].
\[{{f}^{-1}}\left( x \right)=\dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}}\]
Hence, the value of \[{{f}^{-1}}\left( x \right)\] for the function \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\] is \[\dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}}\].
Note: Whenever you get this type of problem, we should always try to make the necessary calculations in the given equation to get the final of x in terms of x which will be the required answer. We should avoid calculation mistakes based on sign conventions. The final solution can also be written as \[{{f}^{-1}}\left( x \right)=\dfrac{1}{\sqrt[3]{x}}\].
We can check the final answer by this method: \[f\left( {{f}^{-1}}\left( x \right) \right)=x\]
Here \[{{f}^{-1}}\left( x \right)=\dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}}\].
Therefore, \[f\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)=\dfrac{1}{{{\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)}^{3}}}\] [since \[f\left( x \right)=\dfrac{1}{{{x}^{3}}}\]]
\[\Rightarrow f\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)=\dfrac{1}{\left( \dfrac{1}{{{x}^{\dfrac{1}{3}\times 3}}} \right)}\]
\[\Rightarrow f\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)=\dfrac{1}{\left( \dfrac{1}{x} \right)}\]
\[\Rightarrow f\left( \dfrac{1}{{{x}^{^{\dfrac{1}{3}}}}} \right)=x\]
\[\therefore f\left( {{f}^{-1}}\left( x \right) \right)=x\]
Hence, the obtained answer is verified.
Recently Updated Pages
Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 7 Maths: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 6 Maths: Engaging Questions & Answers for Success

Class 6 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

