Question

If f(x) and it's inverse function, f^1(x) are both plotted on the same coordinates plane what is their point of intersection

Answer 1

The answer is an equation, a condition:

f(x) = f^(-1)(x), then apply f(x) again: f(f(x) ) = f(f^-1(x)) = x,

f(f(x)) =x, that means that:

The point of intersection is the point where applying f(x) twice it results in the identity. A similar argument takes you to f^(-1)(f^(-1)(x)) = x.

Furthermore, the final answer is the point where f(x)=x (which coincides with f^(-1)(x)=x). That is the value of x where the function crosses the line y=x. If there is no such point, then f(x) and f^(-1)(x) will never cross each other.

I can see the proof graphically, so I can't post it.

For a line, it always works:

f(x) = ax+b, f^(-1)(x) = (x-b)/a,   ax+b = (x-b)/a --> a^2x+ab=x-b,

x = -(a+1)*b/(a^2-1) = -b/(a-1). Which is indeed where f(x)=x.

Answer 2

Answer:

Point of intersection of two graph: It is defined as the point at which where both graph cut to each other.

Step-by-step explanation:

We are given that f(x) and its inverse function is $f^{-1{(x)$ are poltted on the same coordinate plane.

We have to find the point of intersection.

Point of intersection of two graph: It is defined as the point at which where both graph cut to each other and value of both functions are equal at the point.

Suppose a function $f(x)=2x+1$

$f^{-1}(x)=\frac{x-1}{2}$

Substitute x=-1 then we get

$f(-1)=2(-1)-+1=-1$

$f^{-1}(-10)=\frac{-1-1}{2}=-1$

Therefore, the point of intersection is (-1,-1) because the value of both functions are equal at this point.