Let f(x) = (x + 3)²(x - 2)³, x ∈ [-4, 4]. If M and m are the maximum and minimum values of f, respec

Q: Why do we bother checking the endpoints $x = -4$ and $x = 4$? Shouldn't the derivative tell us everything?

This is a classic trap! The derivative only tells us where the slope is zero (local extrema). On a closed interval, the global maximum or minimum can often occur at the boundaries where the function is cut off, even if the slope isn't zero there. Always treat the endpoints as VIP candidates in your search for the absolute extrema.

Q: Is it okay to expand the whole expression before differentiating?

Technically yes, but it is a trap! Expanding $(x + 3)^2(x - 2)^3$ leads to a messy 5th-degree polynomial. Factoring is the mark of a seasoned JEE aspirant. By keeping it in factored form, you save time and reduce the risk of calculation errors—two things that are precious during the exam.

Q: I found $x = 2$ as a critical point, but $f(2) = 0$. Does this mean it's not a maximum or minimum?

It means it is not the *global* maximum or minimum in this specific interval. It is a critical point, but when we compare it against the other candidates (like the endpoints), it turns out to be neither the highest nor the lowest point. Always remember: a critical point is just a candidate; the final verdict comes only after comparing all candidates.

Question

Let $f(x) = (x + 3)^2(x - 2)^3, x \in [-4, 4]$. If $M$ and $m$ are the maximum and minimum values of $f$, respectively in $[-4, 4]$, then the value of $M - m$ is :

Options:

(A) 600

(B) 392

(C) 608

(D) 108

Sigma Percentile · Accepted Answer

### The Landscape of Functions: A Journey into Extrema

Welcome, my dear student. Today, we are not just solving a math problem; we are embarking on a journey to understand the topography of a function.

We are given the function $f(x) = (x + 3)^2(x - 2)^3$ defined on the closed interval $[-4, 4]$. Our mission is to find the absolute maximum, $M$, and the absolute minimum, $m$, and then calculate their difference.

This is a classic JEE Advanced problem that tests not just your ability to differentiate, but your discipline in following the Extreme Value Theorem.

### Phase 1: The Strategy of the Extreme Value Theorem

Before we touch our pens to paper, let us pause. The Extreme Value Theorem tells us that for any continuous function on a closed interval, there must exist an absolute maximum and an absolute minimum.

These values only appear in two types of locations: at the critical points (where the derivative $f'(x)$ is zero or undefined) or at the boundaries of our interval. If you forget the boundaries, you lose the game.

Our roadmap is set: find the derivative, locate the critical points, and then test the entire 'candidate list'—the critical points and the endpoints.

### Phase 2: The Art of Differentiation

Now, let us tackle the derivative. We have $f(x) = (x + 3)^2(x - 2)^3$.

Many students would immediately reach for the expansion, turning this into a massive polynomial. Do not do that! We use the Product Rule: $\frac{d}{dx}[u \cdot v] = u \cdot v' + v \cdot u'$.

Let $u = (x + 3)^2$ and $v = (x - 2)^3$. Applying the rule, we get:

$$f'(x) = (x + 3)^2 \cdot \frac{d}{dx}(x - 2)^3 + (x - 2)^3 \cdot \frac{d}{dx}(x + 3)^2$$

Using the chain rule, this becomes:

$$f'(x) = (x + 3)^2 \cdot 3(x - 2)^2 + (x - 2)^3 \cdot 2(x + 3)$$

Look at this expression. It is already partially factored. Instead of expanding, let us factor out the common terms: $(x + 3)$ and $(x - 2)^2$.

$$f'(x) = (x + 3)(x - 2)^2 [3(x + 3) + 2(x - 2)]$$

Inside the brackets, we simplify: $3x + 9 + 2x - 4 = 5x + 5 = 5(x + 1)$. Thus, our derivative is elegantly reduced to:

$$f'(x) = 5(x + 3)(x - 2)^2(x + 1)$$

### Phase 3: The Hunt for Critical Points

With the derivative in its simplest form, finding the critical points is trivial. We set $f'(x) = 0$.

This gives us three candidates: $x = -3$, $x = 2$, and $x = -1$. All of these lie within our interval $[-4, 4]$.

We have our internal candidates. Now, we must not forget the external candidates: the endpoints $x = -4$ and $x = 4$.

### Phase 4: The Final Tally

Now, we test every candidate. This is the moment of truth.

1. At the left endpoint $x = -4$: $f(-4) = (-4 + 3)^2(-4 - 2)^3 = (-1)^2(-6)^3 = 1 \cdot (-216) = -216$.

2. At the right endpoint $x = 4$: $f(4) = (4 + 3)^2(4 - 2)^3 = (7)^2(2)^3 = 49 \cdot 8 = 392$.

3. At the critical point $x = -3$: $f(-3) = 0$.

4. At the critical point $x = 2$: $f(2) = 0$.

5. At the critical point $x = -1$: $f(-1) = (-1 + 3)^2(-1 - 2)^3 = (2)^2(-3)^3 = 4 \cdot (-27) = -108$.

Our list of values is $\{-216, 392, 0, -108\}$. The maximum value $M$ is clearly $392$, and the minimum value $m$ is $-216$.

### Conclusion: The Final Calculation

Finally, we calculate $M - m$.

$$M - m = 392 - (-216) = 392 + 216 = 608$$

There it is! The **final answer is 608**.

Notice how the entire problem relied on your ability to stay organized and disciplined. We didn't just calculate; we navigated the function's landscape. Keep this mindset, and no problem will ever be too daunting for you.

Animated Solution for Mathematics - Differentiation: Let $f (x) = (x + 3)^{2} (x - 2)^{3}, x \in [- 4, 4]$ . If $M$ and $m$ are the maximum and minimum values of $f$ , respectively in $[- 4, 4]$ , then the value of $M - m$ is :

Select Answer:

Visualized Solution

Analyzing the Function $f (x)$

The Extreme Value Theorem

Applying the Product Rule

Calculating the Derivative

Factoring the Derivative

Locating Critical Points

Testing the Endpoints

Testing the Critical Points

Finding $M$ and $m$

Calculating $M - m$

The Sigma Insight: Maxima and Minima

The Landscape of Functions

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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