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

Question

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

Theodore Amidon · Accepted Answer

#f^(-1)f(x)=x=+-sqrt(f-2)#

The equation #f=x^2+2# represents the parabola, with vertex V(0, 2) and axis along y-axis, in the positive direction, from V. Note that, if (x, f) is on the parabola), so is #(-x, f)#. #xtof# is 2-1 mapping.

So, the inverse obtained by solving for x is #f^(-1)f(x)=x=+-sqrt(f-2)#

The first equation represents the half of the parabola in the 1st quadrant and the second is for the other half, in the second quadrant.

f is single valued but the inverse is double-valued. Inverse trigonometric functions are many-valued, returning a set of values. . .

Sadie Ackerman · Answer

#bar(f)(x)=+-sqrt(x-2)#
#color(white)("XXX")#It is not a function (since it generates more than one solution for single values of #x#.

If #color(red)(bar(f)(x))# is the inverse of #color(blue)(f(x))# then by definition of inverse #color(blue)(f(color(red)(bar(f)(x))))=color(green)(x)# and if #color(blue)(f(color(black)(x)))=color(black)(x)^2+2# then #color(white)("XXX")color(blue)(f(color(red)(bar(f)(x))))=color(red)(bar(f)(x))^2+2# Therefore #color(white)("XXX")color(red)(bar(f)(x))^2+2=color(green)(x)# #color(white)("XXX")color(red)(bar(f)(x))^2=color(green)(x)-2# #color(white)("XXX")color(red)(bar(f)(x))=+-sqrt(color(green)(x)-2)#

Sadie Ackerman · Answer

undefined To find the inverse of $ f(x) = x^2 + 2 $, follow these steps:

1. Replace $ f(x) $ with $ y $.
2. Swap the variables x and y to obtain the equation in terms of y.
3. Solve the resulting equation for y.
4. Replace y with $ f^{-1}(x) $ to express the inverse function.

Starting with $ y = x^2 + 2 $:
1. Swap x and y: $ x = y^2 + 2 $.
2. Solve for y: $ y^2 = x - 2 $ (subtracting 2 from both sides).
3. Taking the square root: $ y = \sqrt{x - 2} $ (or $ y = -\sqrt{x - 2} $ for the negative root).

So, the inverse function is $ f^{-1}(x) = \sqrt{x - 2} $ and $ f^{-1}(x) = -\sqrt{x - 2} $.

To determine if it's a function, we check for one-to-one correspondence. Since $ f(x) = x^2 + 2 $ is a parabola opening upwards, it fails the horizontal line test, meaning it is not one-to-one and therefore does not have an inverse function. However, if we restrict the domain to $ x \geq 2 $ or $ x \leq 2 $, then the inverse is indeed a function.