Home > Precalculus > Introduction to Twelve Basic Functions

How do you determine if #f(x) = ln(x + sqrt(x^2 + 1))# is an even or odd function?

Answer 1

Christian Antunez

#f(x)# is an odd function.

An even function is one for which #f(-x) = f(x)# for all #x# in the domain.

An odd function is one for which #f(-x) = -f(x)# for all #x# in the domain.

Take note of this:

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

Thus, we discover:

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

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

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

So #f(x)# is an odd function.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Greyson Brown

To determine if ( f(x) = \ln(x + \sqrt{x^2 + 1}) ) is an even or odd function, we evaluate ( f(-x) ) and compare it to ( f(x) ). If ( f(-x) = f(x) ), the function is even. If ( f(-x) = -f(x) ), the function is odd.

First, we calculate ( f(-x) ): [ f(-x) = \ln(-x + \sqrt{(-x)^2 + 1}) ] [ = \ln(-x + \sqrt{x^2 + 1}) ]

Next, we compare ( f(-x) ) to ( f(x) ): [ f(-x) \neq f(x) ]

Since ( f(-x) ) is not equal to ( f(x) ), the function is neither even nor odd.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.