a) show that the function f(x)=ln(x+ the square root of (x^2+1)) is an odd function b)find the inverse function of f

Question

a) show that the function f(x)=ln(x+ the square root of (x^2+1)) is an odd function

 

b)find the inverse function of f

Answer 1

(a)

 

For a function to be odd, it must satisfy the property

 

f(-x) = -f(x).

 

f(x) = ln[x + √(x² + 1)]

 

-f(x) = -ln[x + √(x² + 1)]

 

f(-x) = ln{-x + √[(-x)² + 1]} = ln[-x + √(x² + 1)]

 

Now compare f(-x) to -f(x).  Are they equal?

 

 Is ln[-x + √(x² + 1)] = -ln[x + √(x² + 1)] ?

 

Take e⁽⁾ of both expressions and compare them.

 

Is

 

-x + √(x² + 1) = 1 / [x + √(x_² + 1)] ?

 

If they are equal, then 

 

[√(x² + 1) - x] [√(x² + 1) + x] = 1

 

Note you have the factored form of a difference of squares.

 

[√(x² + 1) - x] [√(x² + 1) + x] = [√(x² + 1)]² - x² = (x² + 1) - x² = 1

 

The condition is met, so the function is odd.

 

(b)

 

The get the inverse, exchange x and y and try to express y in terms of x.

 

x = ln[y + √(y² + 1)]

 

e^x = y + √(y² + 1)

 

e^x - y = √(y² + 1)

 

Square both sides.

 

e^2x - 2e^xy + y² = y² + 1

 

e^2x - 2e^xy = 1

 

y = (1 - e^2x) / (-2e^x)

 

f^-1(x) = (e^2x - 1) / (2e^x)

Answer 2

The function f(x) =  ln(x + sqrt(x² +1))  has another name. It is arsinh(x) or  sinh^-1(x)  that is,  the inverse of

the hyperbolic sine function :  sinh(x) = (exp(x) - exp(-x) )/2 

So the answer to part b) is  sinh(x)

 

It is easy to show that sinh(x) is an odd function.  { sinh(-x) =  - sinh(x)  }

 

Since sinh is an odd function, its inverse arsinh(x)  must also be an odd function.   This is the answer to part a)