Let f:[0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If th

Q: Why do we define $g(x) = f(x) - (2x + 3)$? Can't we just look at $f(x)$ directly?

Great question! While you could try to analyze $f(x)$ directly, it is much harder because the line $y = 2x + 3$ is not the x-axis. By defining $g(x)$, we 'flatten' the line. Now, the intersection points of $f(x)$ and the line become the roots of $g(x)$. It turns a complex intersection problem into a simple root-finding problem, which is exactly what Rolle's Theorem is designed to solve.

Q: Wait, why do we count $x=0$ and $x=1$ as roots? The problem says the intersections are in $(0, 1)$.

You are absolutely right that the intersections are in $(0, 1)$. However, the boundary conditions $f(0)=3$ and $f(1)=5$ mean that at $x=0$, $f(0) = 2(0)+3 = 3$, and at $x=1$, $f(1) = 2(1)+3 = 5$. So, $g(0) = 0$ and $g(1) = 0$. Even though they aren't 'intersections' in the sense of crossing, they are still roots of $g(x)$. We must include them to get the full picture!

Q: Does the 'twice differentiable' condition matter? Why not just 'differentiable'?

It is crucial! Rolle's Theorem requires the function to be differentiable to apply it to the first derivative. To apply it to the second derivative, the first derivative must also be differentiable, which means the original function must be twice differentiable. Without this, we couldn't guarantee the existence of $f''(x)$.

Question

Let $f:[0, 1] 	o \mathbb{R}$ be a twice differentiable function in $(0, 1)$ such that $f(0) = 3$ and $f(1) = 5$. If the line $y = 2x + 3$ intersects the graph of $f$ at only two distinct points in $(0, 1)$, then the least number of points $x \in (0, 1)$, at which $f''(x) = 0$, is ________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have a red curve, $f(x)$, that starts at the point $(0, 3)$ and ends at $(1, 5)$. You also have a blue line, $y = 2x + 3$, which connects these exact same two points.

The problem states that the red curve weaves in and out, intersecting our blue line at exactly two distinct points in the open interval $(0, 1)$. This is a beautiful geometric setup, but we can extract the behavior of the second derivative, $f''(x)$, using the elegant power of **Rolle's Theorem**.

### The Auxiliary Function: Our Secret Weapon

When we face a problem involving intersections, we simplify our world by defining an auxiliary function:
$$g(x) = f(x) - (2x + 3)$$

Think of $g(x)$ as the "vertical gap" between the red curve and the blue line. When $g(x) = 0$, the curve and the line are at the same height, meaning they intersect.

Let's examine the boundaries and the given intersections. At $x = 0$, $g(0) = f(0) - (2(0) + 3) = 3 - 3 = 0$. At $x = 1$, $g(1) = f(1) - (2(1) + 3) = 5 - 5 = 0$.

Given the two intersection points $x_1$ and $x_2$ in the interval $(0, 1)$, we have $g(x_1) = 0$ and $g(x_2) = 0$. Consequently, our function $g(x)$ has at least four roots in the interval $[0, 1]$, specifically at $0, x_1, x_2,$ and $1$.

### The Rolle's Cascade

**Rolle's Theorem** is the heartbeat of this solution. It states that if a function has roots at $a$ and $b$, its derivative must have at least one root in $(a, b)$.

Since $g(x)$ has four roots, we apply Rolle's Theorem between each consecutive pair of roots:
1. Between $0$ and $x_1$, there is a root of $g'(x)$.
2. Between $x_1$ and $x_2$, there is a second root of $g'(x)$.
3. Between $x_2$ and $1$, there is a third root of $g'(x)$.

Thus, $g'(x)$ must have at least $3$ roots in the interval $(0, 1)$.

### The Final Revelation

We are looking for $f''(x)$, which is directly related to $g''(x)$. Differentiating our auxiliary function, we get:
$$g'(x) = f'(x) - 2$$

Since $g'(x)$ has at least $3$ roots, we apply Rolle's Theorem again, this time to $g'(x)$. Between any two roots of $g'(x)$, there must be at least one root of its derivative, $g''(x)$.

With $3$ roots for $g'(x)$, we are guaranteed at least $3 - 1 = 2$ roots for $g''(x)$. Because $g''(x) = f''(x)$, the roots of $g''(x)$ are exactly the roots of $f''(x)$.

We have proven that $f''(x)$ must be zero at least twice in the interval $(0, 1)$. Therefore, the **least number of points where $f''(x) = 0$ is 2**.

Enter Numerical Value:

Visualized Solution

Visualizing the Function and the Line

Intersections in the Interval $(0, 1)$

Defining the Auxiliary Function $g (x)$

Finding the Roots of $g (x)$

Total Roots of $g (x)$

Applying Rolle's Theorem to $g (x)$

Geometric Meaning of $g^{'} (x) = 0$

Applying Rolle's Theorem to $g^{'} (x)$

Relating $g^{''} (x)$ to $f^{''} (x)$

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Auxiliary Function

The Rolle's Cascade

The Final Revelation

Similar Questions

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For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$

Let $f : (a, b) \to R$ be twice differentiable function such that $f (x) = \int_{a}^{x} g (t) d t$ for a differentiable function $g (x)$ . If $f (x) = 0$ has exactly five distinct roots in $(a, b)$ , then $g (x) g^{'} (x) = 0$ has at least :

For all $x$ in $[0, 1]$ , let the second derivative $f^{''} (x)$ of a function $f (x)$ exist and satisfy $∣ f^{''} (x) ∣ < 1$ . If $f (0) = f (1)$ , then show that $∣ f^{'} (x) ∣ < 1$ for all $x$ in $[0, 1]$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Function and the Line

Intersections in the Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,1)

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

Finding the Roots of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Total Roots of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Applying Rolle's Theorem to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Geometric Meaning of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(x)=0

Applying Rolle's Theorem to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(x)

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′′(x) to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)