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

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,

& \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{aligned}$$ $$\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{aligned} & \Rightarrow {{f}^{-1}}\left( 0 \right)=0 \\\ & \Rightarrow g\left( 0 \right)=0 \\\ \end{aligned}$$ (b) Here, we have to find the value of g’ (0). Now, from equation (1), we have, $$\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{aligned} & \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{aligned}$$ Substituting x = 0, we have, $$\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{aligned} & \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{aligned}$$ (c) Here, we have to find the value of g (-36). Now, we have, $$\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{aligned} & \Rightarrow -36=f\left( x \right) \\\ & \Rightarrow -36=-35x-{{x}^{5}} \\\ & \Rightarrow {{x}^{5}}+35x=36 \\\ \end{aligned}$$ Clearly, at x = 1 we have the above relation satisfied, that means for x = 1, $$f\left( x \right)=-36$$. $$\begin{aligned} & \Rightarrow {{f}^{-1}}\left( -36 \right)=1 \\\ & \Rightarrow g\left( -36 \right)=1 \\\ \end{aligned}$$ (d) Here, we have to find the value of $$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{aligned} & \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{aligned}$$ **Note:** One must deduce the relation: - $$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)}$$.