Composition with an inverse and intersections of a function and its inverse

Which method is valid for a general one-to-one function and its inverse?

Using inverse composition and intersections

Composing a function with its inverse returns the original input on the appropriate domain. For a general intersection, solve f(x)=f⁻¹(x) on the common input domain. The shortcut f(x)=x requires a separate justification, such as strict increase

1. 1Use the identity on a valid domain A function followed by its inverse returns the original input when that input is in the appropriate domain \(\displaystyle f\left(f^{-1}(x)\right)=x,\quad f^{-1}\left(f(x)\right)=x\)
2. 2Keep the composition order Apply the inner function first, then apply the outer function shown by the composition notation \(\displaystyle f^{-1}\left(f(3)\right)=3\)
3. 3Use a valid common-point equation In general, solve f(x)=f⁻¹(x) on the common input domain, or solve f(f(x))=x with both x and f(x) in the domain of f. Strictly increasing functions have only diagonal intersections; decreasing functions need a separate argument \(\displaystyle y=f(x),\quad x=f(y),\quad x,y\in D_f\)
4. 4Check the declared domain For a general common point, both coordinates must belong to the declared domain of f. Equivalently, the direct equation uses the intersection of the domains of f and f⁻¹. A fixed point in the domain of f automatically passes both checks \(\displaystyle x=-1\notin[0,3]\quad\Rightarrow\quad x=-1\text{ is rejected}\)

Rule

\(\displaystyle f(x)=2x+1\)

For f(x)=2x+1, find the input produced by applying the inverse to 7.

Rule

\(\displaystyle f(x)=2x+1\)

Evaluate the composition shown below using the given rule

Which claim about inverse intersections is justified?

A common point can be part of a two-cycle

At a common point (x,y), the original rule gives y=f(x) and the inverse relation gives x=f(y). Therefore both x and y must lie in the domain of f. This is a two-cycle: f(f(x))=x, with x and f(x) in that domain. A fixed point has x=y, but a two-cycle can use two different numbers. Solving f(x)=f⁻¹(x) directly is valid; it must never be marked wrong merely for being less convenient

Why strict increase justifies the fixed-point shortcut

Suppose f is strictly increasing and (x,y) belongs to both graphs. If x<y, then f(x)<f(y), but the common-point equations give y<x, a contradiction. The case x>y gives the opposite contradiction. Hence x=y. For f(x)=2x+1 on [−2,2], direct substitution gives 2x+1=(x−1)/2, so 3x=−3. The only common point is (−1,−1). Solving f(x)=x gives the same result because the increasing-function argument applies

Decreasing does not mean diagonal-only

Let f(x)=1−x on [−1,2]. Its range is [−1,2], and solving y=1−x for x gives the identical inverse rule. Thus the complete intersection set is {(x,1−x): −1≤x≤2}, not just (1/2,1/2). For example, (0,1) and (1,0) are two different common points. Similarly, the restricted reciprocal function shown above is its own inverse, and both graphs share the whole arc, including (1,2) and (2,1). Its diagonal fixed point (√2,√2) is only one of infinitely many common points

Rule and domain

\(\displaystyle f(x)=1-x,\quad -1\le x\le2\)

For f(x)=1−x on [−1,2], the common point with x=0 is off the diagonal. Find its y-coordinate

Find the full intersection set, not only fixed points

On paper, let q(x)=13/16−x² on [0,1]. First find its range and the positive-root inverse rule. Form q(q(x))=x, factor it using the identity on this screen, and test every candidate against both domain conditions. Give all common points, pairing each x with y=q(x). Explain why the two off-diagonal points are reflections of one another, and why the negative fixed-point root is inadmissible. Finally compare this finite intersection set with the whole-segment self-inverse example

1. 1Do not discard a non-fixed factor The first quadratic gives fixed points. The second gives possible non-diagonal two-cycles. Both must be considered before the original domain restrictions are applied

Three common points, only one diagonal

The endpoints give the range. Rearrange the rule for the inverse, taking the positive root because the original input is nonnegative. The first factor gives fixed points; only the positive root is in the domain. The second factor gives a pair of distinct points whose coordinates both pass the domain checks. These are all the real candidates from the factored equation. This decreasing quadratic has three common points with its inverse, although it is not its own inverse

1. 1Range and inverse Evaluate the rule at the endpoints, then rearrange for the nonnegative input \(\displaystyle q(0)=\frac{13}{16},\quad q(1)=-\frac3{16},\quad -\frac3{16}\le y\le\frac{13}{16},\quad y=\frac{13}{16}-x^2\Rightarrow x=\sqrt{\frac{13}{16}-y}\)
2. 2Fixed point Solve the first factor and reject the root outside the domain \(\displaystyle x^2+x-\frac{13}{16}=0\Rightarrow x=\frac{-2\pm\sqrt{17}}4,\quad a=\frac{\sqrt{17}-2}{4}\in[0,1]\)
3. 3Two-cycle Factor the second quadratic and check both substitutions \(\displaystyle q(q(x))-x=-\left(x^2+x-\frac{13}{16}\right)\left(x^2-x+\frac3{16}\right),\quad x^2-x+\frac3{16}=\left(x-\frac14\right)\left(x-\frac34\right),\quad q\left(\frac14\right)=\frac34,\quad q\left(\frac34\right)=\frac14\)
4. 4Common points The fixed point and the reflected pair give the complete set \(\displaystyle \left(\frac14,\frac34\right),\quad(a,a),\quad\left(\frac34,\frac14\right)\)