If the function f : [0, 4] → R is differentiable then show that (i) For a, b ∈ (0, 4), (f(4))² - (f(

Q: Why can't we just use the Fundamental Theorem of Calculus to solve this?

The Fundamental Theorem of Calculus relates the integral of a derivative to the function values. Here, we are dealing with a product of a derivative and a function value, not an integral. We need to construct the relationship using the properties of continuous and differentiable functions directly, which is why we turn to LMVT and IVT.

Q: How do we know that $a$ and $b$ are distinct points in the interval?

The problem does not require $a$ and $b$ to be distinct. They might be the same point, or they might be different. The theorems guarantee their existence independently within the interval $(0, 4)$, and that is sufficient for the proof to hold regardless of whether they coincide.

Q: Why is the interval $[0, 4]$ so specific in this problem?

The interval $[0, 4]$ is chosen to make the algebra work out cleanly. Specifically, the length of the interval is $4 - 0 = 4$. When we apply the Mean Value Theorem, this $4$ appears in the denominator, which perfectly cancels with the $8$ in the target expression when combined with the factor of $1/2$ from the Intermediate Value Theorem.

Question

If the function $f : [0, 4] 	o \mathbb{R}$ is differentiable then show that (i) For $a, b \in (0, 4), (f(4))^2 - (f(0))^2 = 8f'(a)f(b)$

Sigma Percentile · Accepted Answer

### The Symphony of Analysis: Unlocking the Hidden Geometry

Welcome, aspiring engineers. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of real analysis.

When you look at the expression $(f(4))^2 - (f(0))^2 = 8f'(a)f(b)$, it might seem like a chaotic jumble of variables. But I want you to pause and look closer.

Do you see the structure? It is a classic difference of squares waiting to be revealed. In the world of JEE Advanced, the most complex-looking problems are often just simple truths wearing a disguise.

### Phase 1: The Algebraic Intuition

Before we touch a single theorem, let us look at the target. We want to prove that $(f(4))^2 - (f(0))^2 = 8f'(a)f(b)$.

Recall the algebraic identity: $x^2 - y^2 = (x-y)(x+y)$. If we can show that $8f'(a)f(b)$ is equivalent to $(f(4) - f(0))(f(4) + f(0))$, the proof is essentially done.

This realization is your North Star. It tells us exactly what we need to construct: a term involving the difference of the function values and a term involving the sum of the function values.

### Phase 2: The Titans of Calculus

To build these terms, we need two titans of calculus: the Lagrange Mean Value Theorem (LMVT) and the Intermediate Value Theorem (IVT).

First, let us summon LMVT. We are given that $f$ is differentiable on $[0, 4]$. This implies continuity.

LMVT tells us that there exists some point $a \in (0, 4)$ such that the instantaneous slope $f'(a)$ is equal to the average slope of the secant line connecting the endpoints. Mathematically, this is expressed as:

$$f'(a) = \frac{f(4) - f(0)}{4 - 0} = \frac{f(4) - f(0)}{4}$$

This gives us our first piece of the puzzle: the difference term, $f(4) - f(0)$.

Next, we need the sum term, $f(4) + f(0)$. This is where the Intermediate Value Theorem (IVT) shines.

IVT guarantees that a continuous function takes on every value between its endpoints. Consider the value $\frac{f(4) + f(0)}{2}$. This is the arithmetic mean of the endpoint heights.

Since $f$ is continuous, there must exist some point $b \in (0, 4)$ such that:

$$f(b) = \frac{f(4) + f(0)}{2}$$

### Phase 3: The Synthesis

Now, we have our two components. We have $f'(a)$ and we have $f(b)$. Let us multiply them together, just as the problem demands.

We take our expression $8f'(a)f(b)$ and substitute our findings:

$$8f'(a)f(b) = 8 \cdot \left( \frac{f(4) - f(0)}{4} \right) \cdot \left( \frac{f(4) + f(0)}{2} \right)$$

Look at the constants. We have an $8$ in the numerator. In the denominator, we have $4 \cdot 2$, which is also $8$.

The universe of mathematics is aligning for us. The constants cancel out perfectly, leaving us with:

$$8f'(a)f(b) = (f(4) - f(0))(f(4) + f(0))$$

### Phase 4: The Elegant Conclusion

We have arrived at the final step. By applying the difference of squares identity, we see that $(f(4) - f(0))(f(4) + f(0))$ is exactly $(f(4))^2 - (f(0))^2$.

We have successfully bridged the gap between the derivative at an unknown point and the function values at the boundaries.

This problem teaches us a vital lesson for your JEE preparation: when you see a proof involving derivatives and function values, do not panic. Break the expression down into its algebraic components, and use the theorems of calculus to build those components one by one.

You have the tools; you just need to see the structure. Keep practicing, keep visualizing, and the beauty of calculus will always be within your reach.

Animated Solution for Mathematics - Differentiation: If the function $f : [0, 4] \to R$ is differentiable then show that (i) For $a, b \in (0, 4), (f (4))^{2} - (f (0))^{2} = 8 f^{'} (a) f (b)$

Visualized Solution

Visualizing the Function $f (x)$

Lagrange's Mean Value Theorem

Applying LMVT on $[0, 4]$

Intermediate Value Theorem

Applying IVT on $[0, 4]$

Combining the Results

Simplifying the Constants

Final Algebraic Identity

Key Takeaways

The Sigma Insight: Mean Value Theorems

The Symphony of Analysis

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f:[0,4]→R is differentiable then show that (i) For a,b∈(0,4),(f(4))2−(f(0))2=8f′(a)f(b)