Bijective Crossovers

Finding intersections between $f(x)$ and $f^{-1}(x)$ is similar to finding the number of values of $x$ such that $f(x) = f^{-1}(x)$ , or $f(f(x)) = x$ .

Case 1: $f(x) =x$ for some $x$ , then $f(f(x)) = f(x) = x$ . This implies $x^3+ \frac 14 x -\frac 14 =x$ , or $x^3- \frac 34 x - \frac 14 = 0$ , which can also be written as $\frac 14 (x-1)(2x+1)^2 = 0$ . This equation has two real solutions, namely $-\frac {1} {2}, 1$ .

Case 2: Now, let for some $x$ , $f(x) = k \neq x$ , then, as $f$ is monotonically increasing, if $k>x$ , then $f(k) \geq f(x) > x$ , or $f(f(x)) > x$ , again if $k<x$ , then $f(k) \leq f(x) < x$ , or $f(f(x)) < x$ . So the condition $f(x) = x$ is necessary and sufficient for $f(f(x)) = x$ being true.

Thus, there are only 2 points of intersections.

Due to the symmetry of $f(x)$ and $f^{-1}(x)$ about the line $y=x$ , we need only to see where the curves $f(x)$ and $y=x$ intersect. Put differently, we must find the roots of $g(x)=x^3-\frac{3}{4}x-\frac{1}{4}$ .

Immediately we see that $x=1$ is a solution to $g(x)$ . Polynomial long division leaves us with $g(x)=(x-1)(x^2+x+\frac{1}{4})$ . Note that the discriminant of $(x^2+x+\frac{1}{4})$ , $1-(4)(1)(\frac{1}{4})$ , equals zero so $(x^2+x+\frac{1}{4})$ has only one solution. Using the quadratic formula, we obtain $x=-\frac{1}{2}$ .

We can now completely factor $g(x)=x^3-\frac{3}{4}x-\frac{1}{4}$ into $g(x)=(x-1)(x+\frac{1}{2})^2$ . It is very easy to see that $g(x)$ has two roots, and thus $f(x)$ and $f^{-1}(x)$ intersect exactly twice, at $(1,1)$ and $(-\frac{1}{2},-\frac{1}{2})$

firstly draw the graph of f(x) then draw the graph of f inverse x by taking symmetry along the line x=y then using both the graphs there will be two points of intesection

Finding intersections between $f(x)$ and $f^{-1}(x)$ is similar to finding the number of values of $x$ such that $f(x) = f^{-1}(x)$ , or $f(f(x)) = x$ .

Case 1: $f(x) =x$ for some $x$ , then $f(f(x)) = f(x) = x$ . This implies $x^3+ \frac 14 x -\frac 14 =x$ , or $x^3- \frac 34 x - \frac 14 = 0$ , which can also be written as $\frac 14 (x-1)(2x+1)^2 = 0$ . This equation has two real solutions, namely $-\frac {1} {2}, 1$ .

Case 2: Now, let for some $x$ , $f(x) = k \neq x$ , then, as $f$ is monotonically increasing, if $k>x$ , then $f(k) \geq f(x) > x$ , or $f(f(x)) > x$ , again if $k<x$ , then $f(k) \leq f(x) < x$ , or $f(f(x)) < x$ . So the condition $f(x) = x$ is necessary and sufficient for $f(f(x)) = x$ being true.

Thus, there are only 2 points of intersections.

Note: It is not true that $f(x)$ and $f^{-1} (x)$ only intersect on the line $y = x$ . For example, the inverse of $f(x) = \frac {1}{x}$ is the function itself. If however we know that the function is monotonically increasing, then $f(x)$ and $f^{-1} (x)$ only intersect on the line $y = x$ .

If f is a function such that f:A->B then its inverse f-1(x) is such that f-1(x):B->A For point of intersection, f(x)=f-1(x) so x^3+(1/4)x-(1/4)=x =>4x^3-3x-1=0 =>(x-1)(2x+1)^2=0 =>x=1,x=(-1/2). Therefore two points of intersection.

I began by sketching what the cubic would have to look like to be injective then considered what the inverse function is (a reflection in y = x) this then implied that the roots of f^-1(x) = f(x) were where f(x) = x. By equating, x = x^3 + 1/4 x - 1/4 hence x^3 - 3/4 x - 1/4 = 0 => (x-1)(x+ 1/2)^2 = 0 hence there are two unique points of intersection.

Follow Key technique and then f(x)-x=(x^3+1/4x-1/4)-x=x^3-3/4x-1/4=0 Multiply it by 4 and factorize for x then you'll get (x-1)(2x-1)^2=0 (x-1)=0 has one solution x=1 and (2x-1)^2=0 has one solution x=1/2 Hence function f intersect it's inverse function at two points

If f is invertible (and it is invertible, you found the inverse), then the graph of the inverse function is reflection about the line y=x If they intersect, they will need to intersect on the line y=x So, we can solve the equation

x = f(x)

The given function is one-one. So the oversee function exists. now if f(a)=b then f-1(b)=a. So the f(x)&f-1(x) are symmetric about the line y=x.so f(x) intersects f-1(x) at the line y=x i.e.the solutions of f(x)&f-1(x) is same to that of f(x)=x. The solutions are 1,1,•5.so the function intersect its inverse function at two distinct points 1&•5.

Note that the points of intersection of $y=f(x)$ and $y=f^{-1}(x)$ lie on the line $y=x$ , because every point of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ across $y=x$ . Thus, we are left to finding the number of points of intersection of $y=f(x)=x^3+\frac{1}{4}x-\frac{1}{4}$ and $y=x$ . Substituting, we get the cubic equation $x^3-\frac{3}{4}x-\frac{1}{4}=(x-1)\left(x+\frac{1}{2}\right)^2=0$ . Since there are $2$ real roots, there are $2$ points of intersection.

(might not be right)

You know what a hyperbola looks like.

Then, if you picture it in your head sideways, you can see that there can only be two intersections.

The final answer would have to be two no matter what.

X^3 + 1/4 x-1/4=x The above step is justified because f(x) and f^-1(x) are symmetrical about line y=x. Hence at the intersection points between f^-1(x) and f(x), y=x also intersects the other 2 graphs. Subsequently, we can solve the cubic equation.

X^3-3/4X-1/4=0 4x^3- 3X -1 = 0 By substituting in 1, we can see that x-1 is a factor. (X-1)(4X^2+4X+1)=0 (X-1)(2x+1)^2=0

if f(x) and inverse function meet, then the points must be on the function y=x because f(x) is a monotonically increasing function. this means there will be no points f(x) and inverse function meet which are not on the function y=x. so, solve the equation f(x)=x, then find the number of value. then that's answer :)

they are symmetric by y=x. equation x^3+x/4-1/4=x has 2 roots