You may have learned that we can find the inverse of a function by switching the roles of #x# and #y# and then solve for #y#
For example:
If we have the function #y=x^2+3#
To get it's inverse, we'll switch the roles of #x# and #y# and solve for #y#.
Thus, #y=x^2+3->x=y^2+3#
#x-3=y^2#
#sqrt(x-3)=sqrt(y^2#
#sqrt(x-3)=y# #larr#This is the inverse function denoted by as #f^(-1)(x)#
#--------------------#
Given #sinhx=(e^x-e^-x)/2#, finding it's inverse is the same concept:
Suppose #y=sinhx#, so we rewrite #sinhx=(e^x-e^-x)/2# as...
#y=(e^x-e^-x)/2#
Switch the roles of #x# and #y#
#x=(e^y-e^-y)/2#
Multiply each side by #2#
#color(red)2*x=(e^y-e^-y)/cancel2*cancelcolor(red)2#
#2x=e^y-e^(-y)#
Rewrite #e^-y# as #1/e^y#
#2x=e^y-1/e^y#
Multiply #e^y# to each side
#color(red)(e^y)*2x=(e^y-1/e^y)color(red)(e^y#
#2xe^y=e^(2y)-1#
Subtract #2xe^y# from both sides:
#2xe^ycolor(red)(-2xe^y)=e^(2y)-1-color(red)(2xe^y)#
#0=e^2y-1-2xe^y#
Rewrite as #ax^2+bx+c=0# form
#e^(2y)-2xe^y-1=0#
We must now apply the quadratic formula:
# x = (-b \pm sqrt(b^2-4ac)) / (2a) #
Where #a=1,b=-2x,c=-1# and #x=e^y#
Substituting these values into the quadtratic formula...
# e^y = (-(-2x) \pm sqrt((-2x)^2-4(1)(-1))) / (2(1)) #
# e^y = (2x \pm sqrt(4x^2+4)) / (2) #
Factor #(sqrt(4x^2+4)):#
# e^y = (2x \pm sqrt(4(x^2+1))) / (2) ->(2x \pm sqrt(4)sqrt(x^2+1)) / (2) #
#e^y=(2x \pm 2sqrt(x^2+1)) /2#
Factor #2x \pm 2sqrt(x^2+1)#
#e^y=(2(x \pm sqrt(x^2+1)) )/2#
Simplify
#e^y=(cancelcolor(red)(2)(x \pm sqrt(x^2+1)) )/cancelcolor(red)2#
#e^y=x \pm sqrt(x^2+1)#
Our solution(s) are:
#e^y=x + sqrt(x^2+1),color(red)(cancel(e^y=x-sqrt(x^2+1)#
However, since #y# is real, #e^y# must be positive, and so ignore the second solution: #e^y=x-sqrt(x^2+1)#
So we are left with #e^y=x + sqrt(x^2+1)#
We're not done yet, we still have to solve for #y# but we have an #e# in the way. We can get rid of this #e# quite simply if we take the natural log of both sides since #color(green)(lne^x=x#
Thus, solve for #y#...
#color(red)(ln)e^y=color(red)(ln)(x + sqrt(x^2+1))#
#y=ln(x + sqrt(x^2+1))#
So we can now say, #sinh^(-1)(x)=ln(x + sqrt(x^2+1))#