Animated Solution for Mathematics - Functions: The function $f(x)=\log (x+\sqrt{x^2+1})$, is

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey into the heart of function analysis. We are going to dissect a function that appears frequently in the advanced corridors of calculus: $f(x)=\log (x+\sqrt{x^2+1})$.

This is not just any function; it is the inverse hyperbolic sine, a cornerstone of higher mathematics. Our mission is to determine its parity—to see if it possesses the elegant symmetry of an even function or the rotational grace of an odd function.

Before we touch a single variable, let us ground ourselves in the definitions. Parity is the study of symmetry. We are asking a fundamental question: what happens to the function when we look at it from the other side of the $y$-axis?

Mathematically, we replace $x$ with $-x$.

If $f(-x)=f(x)$, the function is even, meaning it is a perfect mirror image across the $y$-axis. If $f(-x)=-f(x)$, the function is odd, meaning it possesses rotational symmetry of $180^{\circ }$ about the origin. This is our litmus test.

Let us perform the substitution. We replace every $x$ in our function with $-x$. It is crucial here to be meticulous.

Notice how I kept the negative sign inside the parentheses under the square root. We know that any real number squared becomes positive, so $(-x{)}^2$ is simply $x^2$. Our expression simplifies to:

To make this look cleaner, let us rearrange the terms inside the logarithm. Let us put the positive square root first:

Now, look at this expression. It is not immediately obvious how this relates to our original $f(x)=\log (x+\sqrt{x^2+1})$. We have a subtraction inside the logarithm, while our original function has an addition.

This is where we use the 'conjugate' trick. We will multiply and divide the expression inside the logarithm by its conjugate, $(\sqrt{x^2+1}+x)$.

Why do we do this? Because of the difference of squares identity: $(a-b)(a+b)=a^2-b^2$. Here, $a=\sqrt{x^2+1}$ and $b=x$.

When we square $a$, we get $x^2+1$. When we square $b$, we get $x^2$. Subtracting them gives us:

The numerator collapses into a beautiful, simple $1$. Our function now looks like this:

We are almost there. We have a logarithm of a fraction. Recall the fundamental property of logarithms: $\log (\frac{1}{M})=-\log (M)$.

Applying this to our expression, where $M=\sqrt{x^2+1}+x$, we get:

Look closely at the term inside the logarithm. It is exactly our original function $f(x)$!

Therefore, we have proven that:

This is the definition of an odd function. We have successfully navigated the algebra, utilized the conjugate trick, and applied logarithmic properties to reveal the hidden symmetry of the function. It is symmetric about the origin.