Animated Solution for Mathematics - Functions: Let $f:\mathbb{R}\to \mathbb{R}$ and $g:\mathbb{R}\to \mathbb{R}$ be defined as $f(x)=\{\begin{array}{ll}{\log }_{e}x& ,x>0\\ e^{-x}& ,x\le 0\end{array}$ and $g(x)=\{\begin{array}{ll}x& ,x\ge 0\\ e^x& ,x<0\end{array}$. Then, $g\circ f:\mathbb{R}\to \mathbb{R}$ is :

A Journey into $g(f(x))$

Welcome, fellow traveler in the realm of mathematics. Today, we are not just solving a problem; we are deconstructing a machine.

We have two piecewise functions, $f(x)$ and $g(x)$, and we are tasked with understanding their composition, $h(x)=g(f(x))$. Many students fear composite functions, but think of them as a relay race: $f(x)$ runs the first leg, and its output becomes the baton passed to $g(x)$.

The secret to mastering this is to stop looking at $x$ and start looking at the 'state' of the baton.

First, let us define our outer function $g(x)$. It is a gatekeeper defined as:

If the input is non-negative, it lets the value pass through unchanged. If the input is negative, it transforms it exponentially.

Our composite function $g(f(x))$ will behave exactly like this, but the condition depends on the output of $f(x)$. Our goal is to map out the behavior of $f(x)$ across the entire real line.

We must divide the domain into three distinct regions based on how $f(x)$ behaves.

In this region, $f(x)=e^{-x}$. Since $x\le 0$, the exponent $-x$ is non-negative, meaning $f(x)\ge e^0=1$.

Because $f(x)$ is always at least $1$, it is always positive. Looking at our blueprint for $g$, we see that when the input is positive, $g(f(x))=f(x)$.

Thus, in this region:

Now we enter the positive domain where $f(x)={\log }_{e}x$. As $x$ moves from $0$ to $1$, the natural logarithm is negative.

This is the crucial moment! Because $f(x)<0$, we must use the second branch of $g(x)$, which is $g(f(x))=e^{f(x)}$.

Substituting $f(x)$, we get:

By the beautiful laws of logarithms, this simplifies perfectly to:

Finally, for $x\ge 1$, the natural logarithm becomes non-negative. Since $f(x)\ge 0$, we return to the first branch of $g(x)$, where $g(f(x))=f(x)$.

Therefore, in this region:

We have constructed our function:

Now, let us test its properties. Is it one-one?

Draw a horizontal line at $y=0.5$. It hits the line $y=x$ at $x=0.5$ and the curve $y={\log }_{e}x$ at $x=e^{0.5}$. Two inputs for one output confirms that the function is many-one, so it is not one-one.

Is it onto? The range of our function is $[0,\infty )$, which fails to cover the negative real numbers in the co-domain $\mathbb{R}$. Thus, it is not onto.

We have successfully dismantled the problem, piece by piece, and revealed its true nature: it is neither one-one nor onto. Keep this methodical approach in your toolkit, and no composite function will ever intimidate you again.