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

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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.

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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.

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.

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.

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