The number of critical points of the function f(x) = (x - 2)^(2/3)(2x + 1) is

Q: Wait, why do we include $x = 2$ as a critical point if the derivative doesn't exist there?

This is the most common trap in JEE calculus! A critical point is defined as any point $c$ in the domain of $f$ where $f'(c) = 0$ OR $f'(c)$ is undefined. Since $f(2) = (2-2)^{2/3}(2(2)+1) = 0$, the function is perfectly defined at $x = 2$. However, the derivative involves a term in the denominator that becomes zero at $x = 2$. This means the slope becomes infinite—a vertical tangent or a cusp. Because the function exists but the slope does not, it is absolutely a critical point.

Q: Why can't we just use energy conservation or some other shortcut?

Calculus problems like this are testing your fundamental understanding of function behavior, not physical systems. There is no 'shortcut' for finding critical points other than the derivative. You must master the mechanics of differentiation and algebraic simplification. Trust the process—the algebra is the shortcut!

Q: Is it possible to have a critical point that is not a local maximum or minimum?

Absolutely! A critical point is a candidate for a local extremum, but it doesn't guarantee one. For example, $f(x) = x^3$ has a critical point at $x = 0$ (where $f'(0) = 0$), but it is a point of inflection, not a maximum or minimum. In this specific problem, $x=1$ is a local maximum, and $x=2$ is a local minimum. Always check the sign change of the derivative around the critical point to be sure!

Question

The number of critical points of the function $f(x) = (x - 2)^{2/3}(2x + 1)$ is

Options:

(A) 1

(B) 2

(C) 0

(D) 3

Sigma Percentile · Accepted Answer

### The Anatomy of a Function: A Journey into Critical Points

Welcome, future engineers and architects of the future. Today, we are not just solving a math problem; we are performing an autopsy on a function. We are looking for the 'critical points' of $f(x) = (x - 2)^{2/3}(2x + 1)$.

In the world of JEE Advanced, the term 'critical point' is often misunderstood. Students often rush to set the derivative to zero and call it a day. But you, my friends, are going to be more precise. You are going to look for the hidden secrets of the curve.

### Phase 1: The Definition

Before we touch a single variable, let us ground ourselves. What is a critical point? It is a point $x = c$ in the domain of the function where the derivative $f'(c)$ is either zero or undefined.

Think of it geometrically. When $f'(c) = 0$, the curve is momentarily flat—a horizontal tangent. It is the peak of a mountain or the bottom of a valley.

But when $f'(c)$ is undefined, the curve is doing something violent. It is forming a sharp corner, a cusp, or a vertical tangent. These are the points where the function's behavior changes abruptly. If you ignore the 'undefined' case, you are missing half the story.

### Phase 2: The Product Rule Dance

We have $f(x) = (x - 2)^{2/3}(2x + 1)$. This is a product of two functions. Let $u = (x - 2)^{2/3}$ and $v = (2x + 1)$. The Product Rule tells us that $f'(x) = u'v + uv'$.

The derivative of $v = 2x + 1$ is simply $2$. Now, for $u = (x - 2)^{2/3}$, we use the power rule and the chain rule. The $2/3$ comes down, and we subtract $1$ from the exponent, giving us $\frac{2}{3}(x - 2)^{-1/3}$.

So, our derivative is:

$$f'(x) = \frac{2}{3}(x - 2)^{-1/3}(2x + 1) + 2(x - 2)^{2/3}$$

This looks messy, but do not panic. In JEE Advanced, the mess is where the beauty hides.

### Phase 3: The Algebraic Cleanup

We need to simplify this to find our critical points. Let us rewrite the negative exponent as a denominator:

$$f'(x) = \frac{2(2x + 1)}{3(x - 2)^{1/3}} + 2(x - 2)^{2/3}$$

To combine these, we need a common denominator, which is $3(x - 2)^{1/3}$. Multiplying the second term by $\frac{3(x - 2)^{1/3}}{3(x - 2)^{1/3}}$, we get:

$$f'(x) = \frac{2(2x + 1) + 6(x - 2)^{2/3}(x - 2)^{1/3}}{3(x - 2)^{1/3}}$$

Look at the numerator. $(x - 2)^{2/3} \cdot (x - 2)^{1/3}$ is just $(x - 2)^1$. The powers add up to $1$! This is the moment of elegance.

The numerator becomes $2(2x + 1) + 6(x - 2) = 4x + 2 + 6x - 12 = 10x - 10$. Thus, our final, simplified derivative is:

$$f'(x) = \frac{10(x - 1)}{3(x - 2)^{1/3}}$$

### Phase 4: The Discovery

Now, the analysis is trivial. We have two conditions for critical points. First, $f'(x) = 0$. This happens when the numerator is zero: $10(x - 1) = 0$, which gives $x = 1$.

Second, $f'(x)$ is undefined. This happens when the denominator is zero: $3(x - 2)^{1/3} = 0$, which gives $x = 2$.

Both $x = 1$ and $x = 2$ are in the domain of the original function. Therefore, we have exactly two critical points: **$x = 1$ and $x = 2$**.

You see? By staying calm and following the logic, we didn't just find the answer; we understood the geometry of the function. Keep this rigor in your toolkit, and no problem will ever be too daunting. Onward to the next challenge!

Animated Solution for Mathematics - Differentiation: The number of critical points of the function $f (x) = (x - 2)^{2/3} (2 x + 1)$ is

Select Answer:

Visualized Solution

Definition of Critical Points

The Function $f (x)$

Applying the Product Rule

Differentiating $(x - 2)^{\frac{2}{3}}$

Differentiating $(2 x + 1)$

Combining the Derivatives

Finding a Common Denominator

Simplifying the Numerator

The Final Derivative $f^{'} (x)$

Solving $f^{'} (x) = 0$

Identifying Undefined Points

Conclusion: Total Critical Points

The Sigma Insight: Maxima and Minima

The Anatomy of a Function

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

The total number of local maxima and local minima of the function $f (x) = {(2 + x)^{3}, x^{2/3}, - 3 < x \leq - 1 - 1 < x < 2$ is

The function $f (x) = 2 x + 3 (x)^{\frac{2}{3}}, x \in R$ , has

The function $f (x) = 2∣ x ∣ + ∣ x + 2∣ - ∣∣ x + 2∣ - 2∣ x ∣∣$ has a local minimum or a local maximum at $x =$

If $x = 1$ is a critical point of the function $f (x) = (3 x^{2} + a x - 2- a) e^{x}$ , then :

Let $f : R \to R$ be defined as $f (x) = ∣ x ∣ + ∣ x^{2} - 1∣$ . The total number of points at which $f$ attains either a local maximum or a local minimum is

Let $f (x)$ be a polynomial of degree 3 such that $f (- 1) = 10, f (1) = - 6, f (x)$ has a critical point at $x = - 1$ and $f^{'} (x)$ has a critical point at $x = 1$ . Then the local minima at $x = ______$

Let $f (x) = ∣ (x - 1) (x^{2} - 2 x - 3) ∣ + x - 3, x \in R$ . If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ______

Let $f : R \to R$ be a function defined by $f (x) = ∣∣ x + 2∣ - 2∣ x ∣∣$ . If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$ , then $m + n$ is

The range of $a \in R$ for which the function $f (x) = (4 a - 3) (x + lo g_{e} 5) + 2 (a - 7) cot (\frac{x}{2}) sin^{2} (\frac{x}{2})$ , $x \neq = 2 nπ, n \in N$ , has critical points, is

The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of critical points of the function f(x)=(x−2)2/3(2x+1) is