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 larrThis is the inverse function denoted by as f^(-1)(x)
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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))
