Animated Solution for Mathematics - Functions: The function $f:(-\infty ,\infty )\to (-\infty ,1)$, defined by $f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}}$ is:

Welcome, fellow explorer of mathematics! Today, we are going to dissect a function that might look intimidating at first glance, but hides a beautiful, elegant structure.

We are looking at the function defined by:

Our mission is to determine if this function is one-one (injective) and onto (surjective). Let's peel back the layers.

When you see negative exponents like $2^{-x}$, your first instinct should be to simplify. Let's multiply both the numerator and the denominator by $2^x$.

This is a classic algebraic maneuver to clean up the expression:

This is much better! But we can go even further. Let's rewrite the numerator to isolate the variable $x$ more clearly by expressing the numerator as $(2^{2x}+1)-2$.

This allows us to split the fraction:

Now, we have a form that is incredibly easy to analyze. We can see exactly how the function behaves as $x$ changes.

Is the function one-one? To answer this, we need to know if the function is strictly monotonic.

Let's look at the expression $1-\frac{2}{2^{2x}+1}$. As $x$ increases, $2^{2x}$ strictly increases. Consequently, the denominator $2^{2x}+1$ also strictly increases.

If the denominator of a fraction increases, the value of the fraction itself must strictly decrease. Since we are subtracting this decreasing value from $1$, the entire function $f(x)$ must strictly increase.

A strictly increasing function is always one-one because it never returns to a previous $y$-value. It is a one-way street, always moving upward!

Now, for the final piece of the puzzle: is it onto? A function is onto if its range covers the entire codomain.

The problem states the codomain is $(-\infty ,1)$. Let's find the range by looking at the limits:

As $x\to \infty$, $2^{2x}$ grows infinitely large, so $\frac{2}{2^{2x}+1}$ approaches $0$. Thus, $f(x)\to 1-0=1$.

As $x\to -\infty$, $2^{2x}$ approaches $0$, so $\frac{2}{2^{2x}+1}$ approaches $\frac{2}{0+1}=2$. Thus, $f(x)\to 1-2=-1$.

The range of our function is the open interval $(-1,1)$.

We have our results. The function is strictly increasing, so it is one-one.

However, our calculated range is $(-1,1)$, while the given codomain is $(-\infty ,1)$. Since the range is not equal to the codomain, the function is not onto.

We have successfully navigated the trap! The function is one-one but not onto. Keep this analytical mindset, and no function will ever be able to hide its secrets from you again.