Math help? The hyperbolic sine function, sinh x, is defined by the equation:

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2 Answers
Dec 31, 2017

#sinh^-1 (x) = ln(x+sqrt(x^2+1))#

Explanation:

Suppose #y=sinh^-1(x)# so #sinhy =x#

Then #x= (e^y-e^(-y))/2#

# rArr 2x=e^y-e^(-y)#

#rArr2xe^y=e^(2y)-1#

#rArr e^(2y)-2xe^y-1=0#

#rArr e^y = (2x+sqrt(4x^2+4))/2=(2x+2sqrt(x^2+1))/2=x+sqrt(x^2+1)#

#rArr y=sinh^-1(x)=ln(x+sqrt(x^2+1))#

Dec 31, 2017

#sinh^(-1)(x)=ln(x + sqrt(x^2+1))#

Explanation:

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))#