Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:R\to R$ be any function. Define $g:R\to R$ by $g(x)=|f(x)|$ for all $x$. Then $g$ is

Welcome, future engineer! Today, we are going to explore a problem that seems simple on the surface but hides deep conceptual truths about the nature of functions. We are given a function $f:\mathbb{R}\to \mathbb{R}$ and a new function $g(x)=|f(x)|$.

Our goal is to determine which property of $f$ is always inherited by $g$. This is not just about finding the right option; it is about understanding how transformations reshape the landscape of a function.

A Mirror in the Coordinate Plane

Imagine you are standing in front of a mirror. The modulus function $g(x)=|f(x)|$ acts exactly like a mirror placed along the x-axis.

Any part of the graph of $f(x)$ that lies below the x-axis is reflected upwards, becoming positive. This simple geometric act of reflection is powerful, but it comes with consequences.

It can turn a smooth curve into a jagged one, or it can collapse two distinct inputs into the same output. Let us investigate this using the most elegant tool in our arsenal: the counter-example.

When faced with a general function $f(x)$, the best strategy is to test it with a simple, well-behaved function. Let us choose $f(x)=x$.

This function is the gold standard: it is continuous, differentiable, one-one, and onto. Now, let us see what happens when we apply the modulus transformation to it. We get $g(x)=|x|$.

First, let us check if $g(x)=|x|$ is onto. For a function to be onto, its range must equal its codomain, which is $\mathbb{R}$.

But look at the graph of $g(x)=|x|$. It never dips below the x-axis. Its range is $[0,\infty )$.

All the negative real numbers in the codomain are left without a pre-image. Thus, the modulus transformation has destroyed the 'onto' property.

Next, let us check if it is one-one. A function is one-one if every output comes from a unique input. Consider $g(1)=1$ and $g(-1)=1$.

Since two different inputs produce the same output, the function is many-one, not one-one. The mirror effect has collapsed the negative side onto the positive side, destroying injectivity.

Now, let us address the most subtle trap: differentiability. We know $f(x)=x$ is differentiable everywhere.

But look at $g(x)=|x|$ at the origin, $x=0$. There is a sharp 'V' shape, a kink. Mathematically, the left-hand derivative and right-hand derivative are calculated as follows:

Because the slopes from the left and right do not match, the function is not differentiable at $x=0$. The modulus transformation has introduced a point of non-differentiability where none existed before.

Finally, we arrive at continuity. If $f(x)$ is continuous, then $g(x)=|f(x)|$ is the composition of two continuous functions: the function $f(x)$ and the absolute value function $h(u)=|u|$.

Since the absolute value function is continuous everywhere, the composition $h(f(x))$ must also be continuous. There are no jumps, no breaks, and no gaps.

The continuity is preserved. This is the only property that survives the transformation unscathed. We have successfully navigated the traps and arrived at the truth: $g$ is continuous if $f$ is continuous.