Show that the function $f(x)=x^2-x+1$, $x\geq 1/2$ and $\phi(x)=1/2+\sqrt{x-3/4}$ are mutually inverse and solve the equation $x^2-x+1=1/2+\sqrt{x-3/4}$.
This was an example given in the book Problems of Calculus in One Variable written by I.A Maron. The solution gies like this:
The function $y=f(x)=x^2-x+1=(x-1/2)^2+3/4$ increases in the interval $1/2\leq x<\infty$ and with $x$ varrying in the indicated interval, we have $3/4\leq y<\infty$. Hence, defined in the interval $3/4\leq y<\infty$ is the inverse of the function $x=g(y)$ $x\geq 1/2$ , which is found from the equation $x^2-x+(1-y)=0$ . Solving the equation with respect to $x$ , we obtain $x=g(y)=1/2+\sqrt{x-3/4}=\phi(y)$. Let us now solve the equation $x^2-x+1=1/2+\sqrt{x-3/4}$. Since, the graph of the original and inverse functions can intersect only on the straight line $y=x$ , solving the equation $x^2-x+1=x$ , we find $x=1$ .
However, I am not getting how are they concluding "Since, the graph of the original and inverse functions can intersect only on the straight line $y=x$ , solving the equation $x^2-x+1=x$ , we find $x=1$"? How are they saying the graph intersects at the line $y=x$ ? This is an elementary chapter in the book maybe it can be classified as a preliminary chapter "Functions" given before introducing the elementary level concepts of limits of sequences,.... So, maybe this wont require calculus . But I do not get how the conclusion as formerly mentioned is so?
Speaking graphically, inverse of a function $f$ is its reflection about the line $y=x$. So, if the graph of $f$ crosses the line $y=x$, then the inverse of $f$ would cross the line $y=x$ at the same spot/s that $f$ crosses it because of reflection (that should make intuitive sense right?). Hence $f$ and its inverse meet along the line $y=x$.
Speaking mathematically , let $g$ be the inverse of $f$. Since it is claimed that $g$ is the reflection of $f$ about the line $y=x$, we should be able to swap the variables (this swapping is how the said reflection is modeled) in the equation $$ y=g(x) $$ and get the graph of $f$. So, swapping the variables, we get $$ x=g(y)=f^{-1}(y) \implies y=f(x) $$ as required.
Utilizing the same method of swapping for the given inverse function, we get $$ x=1/2+\sqrt{y-3/4} $$ Solving for $y$, we get the graph of the original function $$ y=x^2-x+1 $$
Note:
It is important to note that reflection about the line $y=x$ isn't the inverse of a function, but rather, it is the physical interpretation of the inverse. We can have graphs that are not functions and they'd have reflections about the said line, but are not inverses because the original graph doesn't represent a function to start with.