Let #f(x) = x^2 - 16# how do you find #f^-1(x)#?

Question

16420 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-21T15:04:45

This is a way to express finding the inverse function of #f(x)=x^2-16#

First, write the function as #y=x^2-16#.

Next, switch the #y# and #x# positions.

#x=y^2-16 rarr# Solve for #y# in terms of #x#

#x+16=y^2#

#y=sqrt(x+16)#

The inverse function should be #f^-1(x)=sqrt(x+16)#

Answer 2 · 2018-03-21T17:18:15

Please refer to the Explanation.

Suppose that, # f : RR to RR : f(x)=x^2-16#.

Observe that, #f(1)=1-16=-15, and, f(-1)=-15#.

#:. f(1)=f(-1)#.

#:. f" is not injective, or, "1-1#.

#:. f^-1# does not exist.

However, if # f # is defined on a suitable domain, e.g.,

#RR^+#, then #f^-1# exists as Respected Serena D. has shown.