inverting the function $f:A \rightarrow B$.

Question

Suppose $A$ is the set of real numbers equal to or greater than 2, $B$ is the set of real numbers equal to or greater than 1, and the function $f: A \rightarrow B$ is defined by $f(x)=x^2-4x+5$. How can I find the inverse of the function and the domain and codomain of the inverse function?

The domain shouldn't be difficult - it should just be everywhere that it's defined, and if I differentiate the function I should be able to find critical point(s) and evaluate them with the inverse function to figure out the bounds of the domain.

So the only thing I'm really struggling with is inverting the function $f:A \rightarrow B$.

Answer 1

I'm assuming that you mean $A=\{x\in\mathbb{R}:x\geq 2\}=[2,\infty)$ and $B=\{y\in\mathbb{R}:y\geq 1\}=[1,\infty)$. In other words, $A$ is the set of real numbers greater than or equal to $2$ and $B$ is the set of real numbers greater than or equal to $1$. Instead of $A$ being a real number greater than or equal to $2$, any element of $A$ is a real number greater than or equal to $2$.

Now, consider the function $f(x)=x^2-4x+5$. This defines a parabola and the derivative of this function is $f'(x)=2x-4$. Therefore, the vertex of the parabola is at $x=2$. The value of the function at $x=2$ is $f(2)=1$. Moreover, for $x>2$, the derivative is positive (and increasing). Therefore, the image of $[2,\infty)$ is $[1,\infty)$.

Therefore, we have some hope of inverting the function because the function is onto and one-to-one (one-to-one because its derivative is positive except at $x=2$ and using the mean value theorem).

Starting with $y=f(x)$, you want to find a function so that $f^{-1}(y)=x$. In other words, we start with $$ y=x^2-4x+5 $$ and want to solve for $x$. So, we rewrite this as $$ 0=x^2-4x+(5-y) $$ Using the quadratic formula, we get $$ y=\frac{4\pm\sqrt{16-20+4y}}{2}=\frac{4\pm\sqrt{4y-4}}{2}=2\pm\sqrt{y-1}. $$ Since the codomain $B$ has values greater than or equal to $1$, the square root will never be a square root of a negative number. Now, since $A$ consists of real numbers greater than or equal to $2$, we want the inverse formula $y=2+\sqrt{y-1}$ because the other sign would result in numbers less than $2$.

Then $f^{-1}:B\rightarrow A$ is $f^{-1}(y)=2+\sqrt{y-1}$ and the domain is $B$ and the codomain is $A$.

Answer 2

Hint:

Note that $f(x)=(x-2)^2+1$ and for $y\in[1,\infty)$ find a solution of equation $$(x-2)^2+1=y$$ that lies in $[2,\infty)$.