Function and the inverse-function:
A function f(x) is said to have its inverse say g(x) [ where g(x) = f-1 (x)] , if and only if the f(x) is a one- to one function.
If f(x) and g(x) are inverse functions to each other, then the following conditions are true.
- g(f(x))= x
Horizontal line test:
If a function passes the horizontal line test, then it is a one-to-one function. Else it is not a one-to-one function and we can not have the inverse to the function.
The relation between the derivative of f(x)and the derivative of f-1(x) :
We have the derivative of a function f(x) = f ’(x)
and the derivative of inverse function f-1(x) is represented by [ f-1(x) ]’
Let us derive the relation between these two derivatives.
let us assume that f-1(x)= g(x).
Then as per the inverse properties, we can say that g(f(x)) =x
Let us derivate both sides
Therefore, the derivative of the inverse of a function is 1 over the derivative of the function.
The derivative of an inverse function Worked out examples.
What is the value of the d/dx [ f-1(x)] when x=2,given that f(x)= x3+x
Given that f(x) = x3+x
Therefore the f’ (x) = 3x2+1
Given that the f-1(2)=1 .Therefore f(1) =2.
We have the formula
d/dx [ f-1(x)] when x=2
we initially assumed that the f-1(x) = g(x)
hence d/dx [ f-1(x)] =g’(x).
Therefore we need to find the g’(x) at x=2.
That is the value of g’(2).
From the formula
On comparision we can conclude that f(x) =2
And we can observe that f(x) =x3 +x can be 2, if and only if x=1.
F’(x)= 3x2+1 .Therefore f’(1) = 3(1)2+1 =4
g’(2) = 1/f’(1) = 1/4
Hence, the d/dx [ f-1(x)] when x=2 is 1/4.
The derivative of an inverse function at x=2 is 1/4.