If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6,

Q: Wait, why can't we just use the Mean Value Theorem (MVT) directly on $f$ and $g$?

That is a great question! If you apply MVT to $f(x)$ on $[0, 1]$, you get $f'(c_1) = f(1) - f(0) = 4$. If you apply it to $g(x)$, you get $g'(c_2) = g(1) - g(0) = 2$. While this tells you the average rate of change, it gives you two different points $c_1$ and $c_2$. The problem asks for a single point $c$ where the derivatives are related. By constructing an auxiliary function, we force both functions to 'behave' together at the same point $c$.

Q: How do we know that $\lambda$ must be a constant?

In the context of Rolle's Theorem, we need $h(x)$ to be a well-defined function that satisfies the condition $h(0) = h(1)$. If $\lambda$ were a function of $x$, the derivative $h'(x)$ would involve the product rule and become much more complex, making it impossible to isolate $f'(c)$ and $g'(c)$ in the way the problem requires. By treating $\lambda$ as a constant, we ensure that $h'(x) = f'(x) - \lambda g'(x)$, which leads directly to the linear relationship we need.

Q: Why Rolle's Theorem and not Lagrange's Mean Value Theorem?

Rolle's Theorem is actually a special case of the Mean Value Theorem! When we set $h(0) = h(1)$, we are essentially saying that the average slope of $h(x)$ over the interval $[0, 1]$ is zero. Since the slope is zero, Rolle's Theorem guarantees a point where the instantaneous slope is zero. It is the most direct path to showing that $f'(c) - \lambda g'(c) = 0$.

Question

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f(0) = 2 = g(1), g(0) = 0$ and $f(1) = 6$, then for some $c \in ]0, 1[$

Options:

(A) f'(c) = g'(c)

(B) f'(c) = 2g'(c)

(C) 2f'(c) = g'(c)

(D) 2f'(c) = 3g'(c)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast, uncharted mathematical landscape. You have two paths, represented by the functions $f(x)$ and $g(x)$, both defined on the interval $[0, 1]$.

You know exactly where they start and where they end:
$f(0) = 2$, $f(1) = 6$, $g(0) = 0$, and $g(1) = 2$.

The question before us is profound: is there a hidden connection between the rates at which these two paths change at some intermediate point $c$? This is the essence of **Rolle's Theorem**.

### The Art of the Auxiliary Function

We cannot simply compare $f'(x)$ and $g'(x)$ directly because they are independent entities. We need a bridge.

That bridge is an auxiliary function, $h(x)$. We define:
$$h(x) = f(x) - \lambda g(x)$$
where $\lambda$ is a constant we have yet to discover.

Why this specific form? Because we want to invoke **Rolle's Theorem**, which requires $h(0) = h(1)$. This condition is the key that unlocks the entire problem, as it forces the two functions to synchronize their behavior at the boundaries.

### Solving for the Magic Constant

Let us calculate the values at the boundaries. At $x = 0$, we have:
$$h(0) = f(0) - \lambda g(0) = 2 - \lambda(0) = 2$$

At $x = 1$, we have:
$$h(1) = f(1) - \lambda g(1) = 6 - \lambda(2) = 6 - 2\lambda$$

To satisfy **Rolle's Theorem**, we set $h(0) = h(1)$, which gives us the equation:
$$2 = 6 - 2\lambda$$

Solving this simple linear equation, we find $2\lambda = 4$, which means **$\lambda = 2$**. This is not just a number; it is the scaling factor that aligns the behavior of $f(x)$ and $g(x)$ perfectly.

### The Moment of Truth

With $\lambda = 2$, our auxiliary function becomes $h(x) = f(x) - 2g(x)$. Because $f$ and $g$ are differentiable, $h$ is also differentiable.

Since $h(0) = h(1) = 2$, **Rolle's Theorem** guarantees that there exists at least one point $c \in (0, 1)$ such that $h'(c) = 0$.

Now, we differentiate $h(x)$ with respect to $x$:
$$h'(x) = f'(x) - 2g'(x)$$

Substituting our special point $c$ into this derivative, we get:
$$h'(c) = f'(c) - 2g'(c) = 0$$

Rearranging this, we arrive at the elegant conclusion:
**$$f'(c) = 2g'(c)$$**

We have successfully navigated the terrain, using the boundary conditions to reveal a fundamental relationship between the derivatives of these two functions. This is the power of calculus—taking seemingly disconnected information and finding the hidden harmony beneath.

Animated Solution for Mathematics - Differentiation: If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f (0) = 2 = g (1), g (0) = 0$ and $f (1) = 6$ , then for some $c \in] 0, 1 [$

Select Answer:

Visualized Solution

Analyze Given Conditions

Identifying the Right Theorem

Constructing Auxiliary Function $h (x)$

Evaluating Endpoints for $h (x)$

Finding $λ$

The Final Auxiliary Function

Applying Rolle's Theorem

Differentiating $h (x)$

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Art of the Auxiliary Function

Solving for the Magic Constant

The Moment of Truth

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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 f and g are differentiable functions in [0,1] satisfying f(0)=2=g(1),g(0)=0 and f(1)=6, then for some c∈]0,1[

Select Answer:

Visualized Solution

Analyze Given Conditions

Identifying the Right Theorem

Constructing Auxiliary Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Evaluating Endpoints for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ