If f(x) is a twice differentiable function and given that f(1) = 1; f(2) = 4, f(3) = 9, then (A) f''

Q: Wait, why do we define $g(x) = f(x) - x^2$? Can't we just assume $f(x) = x^2$?

That is a great question! We cannot assume $f(x) = x^2$ because the problem only tells us the values at three specific points. $f(x)$ could be a much more complex function that just happens to pass through $(1, 1)$, $(2, 4)$, and $(3, 9)$. By defining $g(x) = f(x) - x^2$, we create a new function that is guaranteed to be zero at those three points, allowing us to use the power of Rolle's Theorem without making any assumptions about the rest of the curve.

Q: Why does Rolle's Theorem apply to $g'(x)$?

Rolle's Theorem states that if a function is continuous on $[a, b]$, differentiable on $(a, b)$, and $g(a) = g(b)$, then there exists a point $c$ in $(a, b)$ such that $g'(c) = 0$. We found that $g'(x)$ has roots in $(1, 2)$ and $(2, 3)$. Since $g(x)$ is twice differentiable, $g'(x)$ is continuous and differentiable. Therefore, we can apply Rolle's Theorem to $g'(x)$ on the interval between those two roots, which guarantees that $g''(x) = 0$ at some point.

Q: Why is the answer 'for some $x$' and not 'for all $x$'?

Rolle's Theorem is an existence theorem. It guarantees that there is *at least one* point where the derivative is zero. It does not claim that the derivative is zero everywhere. Since we only know the function's values at three points, we cannot conclude anything about the behavior of $f''(x)$ outside of the existence guaranteed by the theorem.

Question

If $f(x)$ is a twice differentiable function and given that $f(1) = 1; f(2) = 4, f(3) = 9$, then

Options:

(A) $f''(x) = 2$ for $\forall x \in (1, 3)$

(B) $f''(x) = f'(x) = 5$ for some $x \in (2, 3)$

(C) $f''(x) = 3$ for $\forall x \in (2, 3)$

(D) $f''(x) = 2$ for some $x \in (1, 3)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at three specific markers: $(1, 1)$, $(2, 4)$, and $(3, 9)$. As you trace these points, a familiar shape emerges—the parabola $y = x^2$.

However, the problem states that $f(x)$ is a twice-differentiable function that passes through these points. It does not necessarily mean $f(x)$ is $x^2$; it could be a complex, oscillating curve that happens to intersect these three targets.

Our goal is to uncover a hidden truth about the second derivative, $f''(x)$, using the elegance of calculus.

### The Difference Engine

To tame this mystery, we introduce an auxiliary function:

$$g(x) = f(x) - x^2$$

Think of $g(x)$ as a "difference engine" that measures the vertical gap between our unknown function $f(x)$ and the standard parabola $y = x^2$.

Since $f(x)$ passes through $(1, 1)$, $(2, 4)$, and $(3, 9)$, we observe the following:

$$g(1) = 1 - 1^2 = 0$$
$$g(2) = 4 - 2^2 = 0$$
$$g(3) = 9 - 3^2 = 0$$

Suddenly, we have a function $g(x)$ that is zero at three distinct locations. This is the breakthrough we need.

### The Rolle's Cascade

Now, we invoke the power of **Rolle's Theorem**. In the interval $[1, 2]$, $g(x)$ starts at zero and ends at zero.

Rolle's Theorem guarantees there is some point $c_1 \in (1, 2)$ where $g'(c_1) = 0$. Similarly, in the interval $[2, 3]$, $g(x)$ also starts at zero and ends at zero, so there must be some point $c_2 \in (2, 3)$ where $g'(c_2) = 0$.

We now have two roots for the derivative function $g'(x)$. If we apply Rolle's Theorem again, this time to $g'(x)$ on the interval $[c_1, c_2]$, we are guaranteed that there exists some $x \in (c_1, c_2)$ such that:

$$g''(x) = 0$$

This is the "cascade"—three roots for $g(x)$ lead to two roots for $g'(x)$, which in turn lead to at least one root for $g''(x)$.

### The Final Revelation

We defined $g(x) = f(x) - x^2$. Differentiating this once gives:

$$g'(x) = f'(x) - 2x$$

Differentiating it a second time gives:

$$g''(x) = f''(x) - 2$$

We just proved that $g''(x) = 0$ for some $x$ in the interval $(1, 3)$. Substituting our expression, we get:

$$f''(x) - 2 = 0$$

This simplifies beautifully to:

$$\mathbf{f''(x) = 2}$$

This is the hidden geometric reality of the problem: no matter how complex $f(x)$ is, if it hits those three points, its second derivative must be $2$ at some point. It is a stunning example of how Rolle's Theorem can extract deep information from minimal data.

Animated Solution for Mathematics - Differentiation: If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

Select Answer:

Visualized Solution

Visualizing the Given Points

Pattern Recognition

Defining the Auxiliary Function

Roots of the Auxiliary Function

Visualizing $g (x)$

Rolle's Theorem (First Interval)

Rolle's Theorem (Second Interval)

Rolle's Theorem on the Derivative

Calculating the Second Derivative

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Difference Engine

The Rolle's Cascade

The Final Revelation

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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(x) is a twice differentiable function and given that f(1)=1;f(2)=4,f(3)=9, then