Animated Solution for Mathematics - Differentiation: Consider the function $f:(0,2)\to \mathbb{R}$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by $g(x)=\{\begin{array}{ll}\min \{f(t)\},& 0<t\le x\text{ and }0<x\le 1\\ \frac{3}{2}+x,& 1<x<2\end{array}$. Then

A Journey into $g(x)$

Welcome, future engineers! Today, we are going to dissect a problem that tests your intuition about functions, monotonicity, and the delicate nature of differentiability.

We are given a function $f(x)=\frac{x}{2}+\frac{2}{x}$ defined on the open interval $(0,2)$. Our mission is to understand the behavior of a new function, $g(x)$, which is defined piecewise.

This is not just about solving for an answer; it is about understanding the 'soul' of the function. Let us begin.

The Monotonicity of $f(x)$

Before we can tackle $g(x)$, we must understand the engine driving it: $f(x)$. To see how $f(x)$ behaves, we look at its rate of change.

We differentiate $f(x)$ with respect to $x$:

This gives us:

Now, look closely at the interval $(0,2)$. For any $x$ in this range, $x^2$ is strictly less than $4$.

This means the numerator $x^2-4$ is always negative. Since the denominator $2x^2$ is always positive, the entire derivative $f^{\prime }(x)$ is negative.

What does this tell us? It tells us that $f(x)$ is a strictly decreasing function. As you walk along the x-axis from $0$ to $2$, the graph of $f(x)$ is constantly sliding downwards.

The Minimum Trap

Now, consider the definition of $g(x)$ for $x\le 1$: $g(x)=\min \{f(t)\}$ for $0<t\le x$.

Because $f(t)$ is strictly decreasing, the smallest value it can take on any interval $(0,x]$ is at the rightmost endpoint, which is $x$. Imagine a slide that goes down; the lowest point you reach is the point where you stop.

Therefore, for $0<x\le 1$:

We have successfully demystified the first piece of our function.

The Junction at $x=1$

We now have a complete definition for $g(x)$:

The critical point is $x=1$. Is the function continuous here? We check the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL).

The LHL is:

The RHL is:

Since the LHL equals the RHL, the function is continuous! There is no jump, no break; the graph is connected.

The Sharp Turn

Finally, we test for differentiability. A function is differentiable only if the slope from the left matches the slope from the right.

The Left-Hand Derivative (LHD) is:

The Right-Hand Derivative (RHD) is the derivative of the second piece:

Since $-1.5 eq 1$, the slopes do not match. There is a sharp corner at $x=1$.

Thus, while the function is continuous, it is not differentiable. You have successfully navigated the trap and uncovered the truth!