Question

Let \[f\left( x \right) = {x^5} + 3x - 2\] and let \[{f^{ - 1}}\] denote the inverse of \[f\]. Then how do you find \[\left( {{f^{ - 1}}} \right)'\left( 2 \right)\] ?

Answer 1

Hint: An inverse function is a function that undoes the action of another function and functions that have inverse are called one to one function. As the question consists of a function in which we need to find the inverse of a function, to solve for the inverse function consider f is a function with inverse function g, and apply chain rule in which we need to find the value of \[g\left( 2 \right)\] and then find its inverse \[g'\left( 2 \right)\] by substituting the value of it in the given function.

Complete step-by-step solution:
A function is said to be one to one if for each number y in the range of f, there is exactly one number x in the domain of f such that \[f\left( x \right) = y\].
For the sake of easier notation, we shall say that f is a function with inverse function g, that is, \[{f^{ - 1}}\left( x \right) = g\left( x \right)\].
According to the definition of inverse functions,
\[\Rightarrow f\left( {g\left( x \right)} \right) = x\]
Differentiating through the chain rule we get
\[\Rightarrow f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right) = 1\]
Solving for the derivative of the inverse gives
\[\Rightarrow g'\left( x \right) = \dfrac{1}{{f'\left( {g\left( x \right)} \right)}}\]
As we need to find \[g'\left( 2 \right)\] i.e.,
\[\Rightarrow g'\left( 2 \right) = \dfrac{1}{{f'\left( {g\left( 2 \right)} \right)}}\] …………………… 1
We want to first find \[g\left( 2 \right)\], however, we cannot write an expression for \[g\left( x \right)\].
Note that if \[f\left( x \right) = 2\], then \[x = g\left( 2 \right)\],
This means we should let \[f\left( x \right) = 2\]then solve for x, which is equal to \[g\left( 2 \right)\].
\[\Rightarrow f\left( x \right) = 2\]
\[ \Rightarrow {x^5} + 3x - 2 = 2\]
Continuing to solve yields
\[\Rightarrow {x^5} + 3x - 4 = 0\]
The sum of the coefficients of each term is 0, that is,
\[\Rightarrow 1 + 3 - 4 = 0\].
 This means that x = 1 is a solution.
Dividing, we can see that
\[\Rightarrow \left( {x - 1} \right)\left( {{x^4} + {x^3} + {x^2} + x + 4} \right) = 0\]
Note that,
 \[\Rightarrow {x^4} + {x^3} + {x^2} + x + 4 > 0\] for all values of x, so it has no real roots.
Thus,
\[\Rightarrow f\left( x \right) = 2\] when, \[x = 1\] i.e.,
\[\Rightarrow f\left( 1 \right) = 2\]
This implies that:
\[\Rightarrow g\left( 2 \right) = 1\]
As from equation 1 we know that,
\[\Rightarrow g'\left( 2 \right) = \dfrac{1}{{f'\left( {g\left( 2 \right)} \right)}}\]
Hence, substitute the value of \[g\left( 2 \right)\] as \[g\left( 2 \right) = 1\]we get
\[\Rightarrow g'\left( 2 \right) = \dfrac{1}{{f'\left( {g\left( 2 \right)} \right)}} = \dfrac{1}{{f'\left( 1 \right)}}\] ………………….. 2
Hence, we can find this by taking the derivative of \[f\] by the given function i.e.,
\[\Rightarrow f\left( x \right) = {x^5} + 3x - 2\]
The derivative of the function \[f\left( x \right)\] is:
\[\Rightarrow f'\left( x \right) = 5{x^4} + 3\]
Thus, now we can find \[f'\left( 1 \right)\],
\[\Rightarrow f'\left( 1 \right) = 5{\left( 1 \right)^4} + 3\]
\[\Rightarrow f'\left( 1 \right) = 5 + 3\]
\[ \Rightarrow f'\left( 1 \right) = 8\]
As we know the value of \[f'\left( 1 \right)\], now let us find the inverse of \[g'\left( 2 \right)\] from equation 2 i.e.,
\[\Rightarrow g'\left( 2 \right) = \dfrac{1}{{f'\left( {g\left( 2 \right)} \right)}} = \dfrac{1}{{f'\left( 1 \right)}}\]
\[\Rightarrow g'\left( 2 \right) = \dfrac{1}{{f'\left( 1 \right)}} = \dfrac{1}{8}\]

Hence, \[\left( {{f^{ - 1}}} \right)'\left( 2 \right)\]= \[\dfrac{1}{8}\]

Note: The key point to find the inverse as we have considered f is a function with inverse function g, this implies that we must look on to the function that the given function is composite function, hence we must apply chain rule to differentiate it easily as the chain rule tells us how to find the derivative of a composite function and it states that the derivative of \[f\left( {g\left( x \right)} \right)\] is \[f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\], hence differentiate and find the inverse of given function.