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

2 Answers
Apr 18, 2016

#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#.

Explanation:

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)#

Apr 18, 2016

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

Explanation:

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. .
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