#f# is a function on #RR# where #f(x)=|x^2-2|#, How do you find #f^-1(x)#?

2 Answers
Feb 5, 2018

Look below

Explanation:

first, we have a detail that the function #f# is present for #RR#.

the function is #\abs{x^2-2}#, so let's make a table from #x >= 0# to #x <= 2#

Paint

now graph the points, #(0, 2), (1, 1), (2, 2)#

graph{\abs{x^2-2}[-10, 10, -5, 5]}

now, the inverse of an absolute value function is basically the same function, which is #abs{ x^2-2}#

Feb 5, 2018

desmos.com

Explanation:

Let's turn #f(x)=abs(x^2-2)# into a piece-wise function.

Now, we see that #absa=a# when #a>=0#
Similarly, we see that #absa=-a# when #a<0#

Therefore, we have:
#f(x)=x^2-2# when #x^2-2>=0#

#f(x)=-(x^2-2)# when #x^2-2<0#

Let's graph the two parabolas. When we do, we get:

desmos.com
Now, the blue parabola applies when #x^2-2>=0#, and green parabola applies when #x^2-2<0#.

Therefore, we have(focus on the red graph):
desmos.com
Now, to find its inverse, we have to reflect the red graph over the line #y=x#.

Let's try this mathematically.

Our piece-wise function was:
#f(x)=x^2-2# when #x^2-2>=0#

#f(x)=-(x^2-2)# when #x^2-2<0#

Let's find the inverse function for each situation.

Now, remember that #f(f^-1(x))=x# and #f^-1(f(x))=x#

Therefore, we can now find the inverse functions.

#f(x)=x^2-2=>x=(f^-1(x))^2-2#

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

=>#+-sqrt (x+2)=f^-1(x)# when #x^2-2>=0#

Similarly,

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

=>#x-2=-(f^-1(x))^2#

=>#2-x=(f^-1(x))^2#

=>#+-sqrt (2-x)=f^-1(x)# when #x^2-2<0#

We now graph these two sideways parabolas:
desmos.com
When #y>=sqrt2# or #y<= -sqrt2 # , then we apply our first function.

When #-sqrt2<y< sqrt2# , then we apply our second function.

We therefore have:

desmos.com

Just note that #+-sqrt(x+2)# and other relations like this one is not a function but a relation because there are more than one output for a given input.