Prove that f is invertible and find (f^-1)'((1)/(2) ?

If #f(x) = cos x# for all x #in# #(0, (pi)/(2))#, prove that f is invertible and find #(f^-1)'((1)/(2))#.

1 Answer
Mar 31, 2018

#arccos(1/2)= pi/3#

Explanation:

To know if a relation is invertible we must know the criteria for being invertible. The criteria are as follows:

❈ It must be a function, meaning that each value of #x# maps to only 1 #y# value.

❈ The inverse function must also be a function, meaning that each value of #x# maps to only 1 #y# value.

❈ The inverse function must be a reflection of #y=f(x)# in the line #y=x#. (you can test this using http://desmos.com/calculator)

❈ Must satisfy #f(f^{-1}(x)) = x# and #f^{-1}(f(x)) = x#

First of all, as #f(x) = cosx# is a function that maps to only 1 value for each x value, it is invertible for at least a certain range.

The inverse function of #f(x)=cosx# is #f^{-1}(x)=arccos(x)#, by definition.

However since #cosx# is a repeating function it is a many-to-one function, meaning that several x values will give the same y value. This means that the inverse function will be a one-to-many relation (since y and x switch places), and therefore not a function. This problem can be overcome by restricting the domain, which has already been done: #x in (0, pi/2)#. The function is defined for this domain.

#arccos(1/2)= pi/3#

Now to prove that it is invertible we want to test it using the following equations:

#f(f^{-1}(x)) = x# and #f^{-1}(f(x)) = x#

#cos(arccos(1/2)) =1/2#
#arccos(cos(pi/3)) = pi/3#