Consider the function defined as f(x) = e^x + e^-x/2, x > 0. Find the inverse of f?

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Answer 1 · 2018-04-18T03:34:24

Assuming #f(x)=(e^x+e^-x)/2#, then

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

But if #f(x)=e^x+e^-x/2#, then

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

I'm not sure if you wrote the function correctly. You wrote

#f(x)=e^x+e^-x/2#

Did you mean to find the inverse of

#f(x)=(e^x+e^-x)/2#?

If so, start with

#f(x)=y=(e^x+e^-x)/2#

Multiply both sides by 2.

#2y=e^x+e^-x#

Multiply both sides by #e^x#.

#2ye^x=(e^x)^2+1#

Subtract #2ye^x# from both sides.

#0=(e^x)^2-2ye^x+1#

Use the quadratic formula to solve for #e^x#.

#e^x=ypmsqrt(y^2-1)#

But #e^x# and #y# are all positive so we can just write

#e^x=y+sqrt(y^2-1)#

Now take the natural log of both sides.

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

Another way to say this is

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

If you really DID mean #f(x)=e^x+e^-x/2#, then the answer is

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

using exactly the same method as above.