Question

Let \[f\left( x \right)=-35x-{{x}^{5}}\] and let g be the inverse function of f, how do you find (a) g (0) (b) g’ (0) (c) g (-36) (d) g’ (-36)?

Answer 1

Hint: Write \[g\left( x \right)={{f}^{-1}}\left( x \right)\] and convert it into the relation \[f\left( g\left( x \right) \right)=x\]. Now, to find g (0) and g (-36), find \[{{f}^{-1}}\left( 0 \right)\] and \[{{f}^{-1}}\left( 36 \right)\] respectively. That means find the value of x for which \[f\left( x \right)=0\] and \[f\left( x \right)=-36\] respectively. In the second part of the question find g’ (x) by differentiating \[f\left( g\left( x \right) \right)=x\] both the sides with respect to x. Calculate the values of g’ (0) and g’ (-36) using the obtained values of g (0) and g (-36) respectively.

Complete step by step answer:
Here, we have been provided with the function \[f\left( x \right)=-35x-{{x}^{5}}\] and it is given that g (x) is the inverse of the function \[f\left( x \right)\]. So, we have,
\[\Rightarrow g\left( x \right)={{f}^{-1}}\left( x \right)\]
\[\Rightarrow f\left( g\left( x \right) \right)=x\] - (1)
(a) Here, we have to find the value of g (0). Now, we have,
\[\Rightarrow g\left( 0 \right)={{f}^{-1}}\left( 0 \right)\]
Now, \[{{f}^{-1}}\left( 0 \right)\] means we have to find such a value of x for which the function \[f\left( x \right)\] equals 0. So, we have,
\[\begin{align}
  & \because f\left( x \right)=0 \\
 & \Rightarrow -35x-{{x}^{5}}=0 \\
 & \Rightarrow -x\left( 35+{{x}^{4}} \right)=0 \\
 & \Rightarrow x\left( {{x}^{4}}+35 \right)=0 \\
\end{align}\]
\[\Rightarrow x=0\], because \[\left( {{x}^{4}}+35 \right)\] will always be positive.
That means for x = 0 we have, \[f\left( x \right)=0\].
\[\begin{align}
  & \Rightarrow {{f}^{-1}}\left( 0 \right)=0 \\
 & \Rightarrow g\left( 0 \right)=0 \\
\end{align}\]
(b) Here, we have to find the value of g’ (0). Now, from equation (1), we have,
\[\Rightarrow f\left( g\left( x \right) \right)=x\]
Differentiating both the sides with respect of x, and using the chain rule of differentiation in the L.H.S., we get,
\[\begin{align}
  & \Rightarrow f'\left( g\left( x \right) \right)\times g'\left( x \right)=1 \\
 & \Rightarrow g'\left( x \right)=\dfrac{1}{f'\left( g\left( x \right) \right)} \\
\end{align}\]
Substituting x = 0, we have,
\[\Rightarrow g'\left( 0 \right)=\dfrac{1}{f'\left( g\left( 0 \right) \right)}\]
Substituting the obtained value of g (0) in part (a), we get,
\[\Rightarrow g'\left( 0 \right)=\dfrac{1}{f'\left( 0 \right)}\]
So, we need to find \[f'\left( x \right)\] at x = 0. Therefore, differentiating the function \[f\left( x \right)\] with respect to x, we get,
\[\begin{align}
  & \Rightarrow f'\left( x \right)=-35-5{{x}^{4}} \\
 & \Rightarrow f'\left( 0 \right)=-35 \\
 & \Rightarrow g'\left( 0 \right)=\dfrac{1}{-35} \\
 & \Rightarrow g'\left( 0 \right)=\dfrac{-1}{35} \\
\end{align}\]
(c) Here, we have to find the value of g (-36). Now, we have,
\[\Rightarrow g\left( -36 \right)={{f}^{-1}}\left( -36 \right)\]
\[\Rightarrow {{f}^{-1}}\left( -36 \right)\] means we have to find the value of x for which the function \[f\left( x \right)\] equals -36. So, we have,
\[\begin{align}
  & \Rightarrow -36=f\left( x \right) \\
 & \Rightarrow -36=-35x-{{x}^{5}} \\
 & \Rightarrow {{x}^{5}}+35x=36 \\
\end{align}\]
Clearly, at x = 1 we have the above relation satisfied, that means for x = 1, \[f\left( x \right)=-36\].
\[\begin{align}
  & \Rightarrow {{f}^{-1}}\left( -36 \right)=1 \\
 & \Rightarrow g\left( -36 \right)=1 \\
\end{align}\]
(d) Here, we have to find the value of \[g'\left( -36 \right)\]. So, from the relation \[g'\left( x \right)=\dfrac{1}{f'\left( g\left( x \right) \right)}\] that we have obtained in part (b), we get,
\[\Rightarrow g'\left( -36 \right)=\dfrac{1}{f'\left( g\left( -36 \right) \right)}\]
Substituting the obtained value of \[g\left( -36 \right)\] in part (c), we get,
\[\Rightarrow g'\left( -36 \right)=\dfrac{1}{f'\left( 1 \right)}\]
So, we need to find \[f'\left( x \right)\] at x = 1. So, we have,
\[\begin{align}
  & \because f'\left( x \right)=-35-5{{x}^{4}} \\
 & \Rightarrow f'\left( x \right)=-35-5 \\
 & \Rightarrow f'\left( 1 \right)=-40 \\
 & \Rightarrow g'\left( -36 \right)=\dfrac{1}{-40} \\
 & \Rightarrow g'\left( -36 \right)=\dfrac{-1}{40} \\
\end{align}\]

Note: One must deduce the relation: - \[f\left( g\left( x \right) \right)=x\] in order to solve the question. Remember the chain rule of differentiation used for the derivative of a composite function. You may note that we solved the equation \[{{x}^{5}}+35x=36\] by the hit and trial method, substituting x = 1. This is because it is an equation of degree 5 and cannot be solved algebraically. Remember that in these types of questions you will not be given harsh calculations but you must remember the formula: - \[g'\left( x \right)=\dfrac{1}{f'\left( g\left( x \right) \right)}\].