If f(x) = x(√x - √(x+1)), then (a) f(x) is continuous but not differentiable at x = 0 (b) f(x) is di

Q: Wait, why do we only check the Right-Hand Derivative? Shouldn't we check the Left-Hand Derivative too?

That is a fantastic question! Differentiability at a point requires the derivative to exist from both sides. However, our function $f(x) = x(\sqrt{x} - \sqrt{x+1})$ is only defined for $x \ge 0$. Since the function doesn't exist for $x < 0$, the Left-Hand Derivative is undefined. In calculus, for a function defined on a closed interval, we only require the one-sided derivative at the boundary point for it to be considered differentiable there.

Q: Why does the derivative exist if the function involves $\sqrt{x}$? Isn't $\sqrt{x}$ non-differentiable at $x=0$?

You have a sharp eye! You are absolutely correct that $\sqrt{x}$ has an infinite slope at $x=0$. However, our function is $f(x) = x\sqrt{x} - x\sqrt{x+1}$. The multiplication by $x$ acts as a 'taming' factor. As $x$ approaches zero, the $x$ term effectively 'crushes' the singularity of the $\sqrt{x}$ term, allowing the derivative to settle into a finite value.

Q: Is it possible for a function to be differentiable at a point where its individual components are not?

Yes, it is! This is a classic JEE concept. While $\sqrt{x}$ is not differentiable at $x=0$, the product $x\sqrt{x}$ is. The extra factor of $x$ provides the necessary smoothness to make the derivative finite. Always look at the function as a whole rather than just its individual parts.

Question

If $f(x) = x(\sqrt{x} - \sqrt{x+1})$, then

Options:

(A) $f(x)$ is continuous but not differentiable at $x = 0$

(B) $f(x)$ is differentiable at $x = 0$

(C) $f(x)$ is not differentiable at $x = 0$

(D) none of these

Sigma Percentile · Accepted Answer

### The Art of Taming Singularities: A Journey into Differentiability at the Boundary

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a problem that often trips up even the most seasoned JEE aspirants.

We are looking at the function $f(x) = x(\sqrt{x} - \sqrt{x+1})$. At first glance, it looks like a standard calculus problem, but it hides a beautiful lesson about the nature of boundaries and the 'taming' power of algebra.

### Phase 1: Respecting the Domain

Before we even touch a derivative, we must respect the domain. The presence of $\sqrt{x}$ acts like a gatekeeper. It tells us that $x$ cannot be negative.

Therefore, our function lives in the world of $x \ge 0$. This is crucial because it changes the definition of differentiability.

We are not looking at an interior point where we can approach from both sides; we are standing at the very edge of the cliff—the boundary point $x=0$.

### Phase 2: The Boundary Logic

Many students panic here, asking, 'What about the Left-Hand Derivative?' But remember, the function simply does not exist for $x < 0$.

In the realm of real analysis, for a function defined on a closed interval like $[0, \infty)$, we only need the Right-Hand Derivative (RHD) to exist for the function to be differentiable at that endpoint.

We are looking for the slope of the tangent as we move away from the origin into the positive territory.

### Phase 3: The First Principle

Let us invoke the First Principle of Derivatives. It is the heartbeat of calculus. We want to find the slope at $x=0$, which is defined as:

$$f'(0^+) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h}$$

We know $f(0) = 0$. Now, let us substitute $h$ into our function. Our expression becomes $f(h) = h(\sqrt{h} - \sqrt{h+1})$.

Plugging this into our limit, we get:

$$f'(0^+) = \lim_{h 	o 0^+} \frac{h(\sqrt{h} - \sqrt{h+1}) - 0}{h}$$

### Phase 4: The Algebraic Taming

Here comes the magic. Look at the $h$ in the numerator and the $h$ in the denominator.

Because we are taking a limit as $h 	o 0^+$, we know $h$ is not exactly zero. This allows us to perform the most satisfying cancellation in algebra.

The $h$ terms vanish, leaving us with a clean, manageable expression:

$$f'(0^+) = \lim_{h 	o 0^+} (\sqrt{h} - \sqrt{h+1})$$

Now, we simply evaluate the limit by letting $h$ approach zero. The first term, $\sqrt{h}$, becomes $\sqrt{0} = 0$. The second term, $\sqrt{h+1}$, becomes $\sqrt{0+1} = 1$.

Thus, our result is $0 - 1 = -1$.

### The Final Revelation

We have arrived at the answer: the Right-Hand Derivative is **$-1$**.

Geometrically, this tells us that at the origin, the curve has a well-defined tangent with a slope of $-1$. The function is not only continuous but perfectly differentiable at the boundary.

This problem teaches us that even when individual components of a function seem problematic, the interaction between them—in this case, the multiplication by $x$—can create a smooth, differentiable path. Keep this in mind: never judge a function by its parts; always look at the whole.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (x) = x (x - x + 1)$ , then

Select Answer:

Visualized Solution

Analyzing the Function

Continuity at $x = 0$

Differentiability at a Boundary Point

The First Principle Formula

Substituting the Function Values

Setting Up the Limit

Simplifying the Expression

Evaluating the Limit

Geometric Interpretation

Final Conclusion

The Sigma Insight: Differentiability of a Function

The Art of Taming Singularities

Phase 1

Phase 2

Phase 3

Phase 4

The Final Revelation

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x(x​−x+1​), then