How do you find the inverse of #h(x)= x^2 - 4x + 5# and is it a function?

Question

How do you find the inverse of  #h(x)= x^2 - 4x + 5# and is it a function?

Sadie Ackerman · Accepted Answer

Inversion gives two forms for #x = 2+-sqrt(h-1), h>=1 #

The for this equation is a parabola

#(x-2)^2=y-1>=0, and# so, #y>=1#,

The parabola is symmetrical about its axis x = 2.

Explicitly, the inverse functions are

#x = 2+-sqrt(h-1), h>=1 #, representing the halves of the parabola,

bifurcated by x = 2.l

Sadie Alexander · Answer

undefined To find the inverse of a function, $ h(x) = x^2 - 4x + 5 $, follow these steps:

1. Replace $ h(x) $ with $ y $.
2. Swap $ x $ and $ y $, making the equation: $ x = y^2 - 4y + 5 $.
3. Rearrange the equation to solve for $ y $.
4. After solving for $ y $, replace $ y $ with $ f^{-1}(x) $, which represents the inverse function.

Whether the inverse function is also a function depends on whether the original function is one-to-one. A function is one-to-one if each input has a unique output. In this case, since the original function $ h(x) = x^2 - 4x + 5 $ is a quadratic, we can determine if it's one-to-one by checking if its graph passes the horizontal line test. If the graph intersects any horizontal line at most once, then the function is one-to-one, and its inverse will also be a function. If the graph intersects a horizontal line more than once, then it's not one-to-one, and its inverse won't be a function.