Key Concepts and Formulas

• Rolle's Theorem: If a function $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
• Generalized Rolle's Theorem (or Mean Value Theorem for higher derivatives): If a function $f$ is $n$ times differentiable on an interval, and if $f(x)$ has $n+1$ distinct roots in that interval, then $f^{(n)}(x)$ has at least one root in the same interval.
• Relationship between $f''(x)$ and Concavity: The sign of the second derivative, $f''(x)$, indicates the concavity of the function.
  • If $f''(x) > 0$, the function is concave up.
  • If $f''(x) < 0$, the function is concave down.
  • If $f''(x) = 0$, the point might be an inflection point where concavity changes.

Step-by-Step Solution

Step 1: Define a helper function. Let's define a new function $g(x) = f(x) - (2x+3)$. This function will help us analyze the points where the graph of $f(x)$ intersects the line $y = 2x+3$. The roots of $g(x)$ correspond to the intersection points of $f(x)$ and $y=2x+3$.

Step 2: Determine the properties of the helper function $g(x)$. Since $f(x)$ is twice differentiable in $(0,1)$, $g(x)$ is also twice differentiable in $(0,1)$. The derivative of $g(x)$ is $g'(x) = f'(x) - 2$. The second derivative of $g(x)$ is $g''(x) = f''(x)$.

Step 3: Use the given information about the intersection points. We are given that the line $y = 2x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$. Let these points be $x_1$ and $x_2$, where $0 < x_1 < x_2 < 1$. At these intersection points, $f(x) = 2x+3$, which means $f(x) - (2x+3) = 0$. Therefore, $g(x_1) = 0$ and $g(x_2) = 0$.

Step 4: Apply Rolle's Theorem to $g(x)$. Since $g(x)$ is continuous on $[x_1, x_2]$ (as $f$ is continuous on $[0,1]$) and differentiable on $(x_1, x_2)$, and $g(x_1) = g(x_2) = 0$, by Rolle's Theorem, there exists at least one point $c_1 \in (x_1, x_2)$ such that $g'(c_1) = 0$. Substituting back $g'(x) = f'(x) - 2$, we get $f'(c_1) - 2 = 0$, which means $f'(c_1) = 2$.

Step 5: Consider the boundary conditions for $f(x)$. We are given $f(0)=3$ and $f(1)=5$. Let's evaluate $g(x)$ at the boundaries of the interval $[0,1]$: $g(0) = f(0) - (2(0)+3) = 3 - 3 = 0$. $g(1) = f(1) - (2(1)+3) = 5 - 5 = 0$.

Step 6: Apply Rolle's Theorem to $g(x)$ on the interval $[0,1]$. Since $g(x)$ is continuous on $[0,1]$, differentiable on $(0,1)$, and $g(0) = g(1) = 0$, by Rolle's Theorem, there exists at least one point $c_0 \in (0,1)$ such that $g'(c_0) = 0$. This implies $f'(c_0) = 2$.

Step 7: Apply Rolle's Theorem to $g'(x)$. We know that $g'(x)$ is differentiable on $(0,1)$ (since $f''(x)$ exists). We have found that $g'(x) = 0$ at at least two distinct points in $(0,1)$: $c_0$ and $c_1$, where $0 < c_0 < 1$ and $x_1 < c_1 < x_2$. Since $x_1$ and $x_2$ are distinct points in $(0,1)$, and $c_1$ is strictly between them, $c_1$ is also in $(0,1)$. Also, we know $g(0) = 0$ and $g(x_1) = 0$. Applying Rolle's theorem on $[0, x_1]$ for $g(x)$, there exists $d_1 \in (0, x_1)$ such that $g'(d_1) = 0$. We know $g(x_1) = 0$ and $g(1) = 0$. Applying Rolle's theorem on $[x_1, 1]$ for $g(x)$, there exists $d_2 \in (x_1, 1)$ such that $g'(d_2) = 0$. So we have at least three distinct points where $g'(x) = 0$: $d_1 \in (0, x_1)$, $d_2 \in (x_1, 1)$. Let's re-evaluate the roots of $g(x)$. We are given that $y=2x+3$ intersects $f(x)$ at only two distinct points in $(0,1)$. Let these points be $a$ and $b$, with $0 < a < b < 1$. So, $g(a) = 0$ and $g(b) = 0$.

Now consider the values at the endpoints of the interval $[0,1]$: $g(0) = f(0) - (2(0)+3) = 3 - 3 = 0$. $g(1) = f(1) - (2(1)+3) = 5 - 5 = 0$.

So, we have $g(0) = 0$, $g(a) = 0$, $g(b) = 0$, $g(1) = 0$. These are four distinct points in $[0,1]$ where $g(x) = 0$. Since $0 < a < b < 1$, the four points are $0, a, b, 1$.

Applying Rolle's Theorem to $g(x)$ on the intervals $[0, a]$, $[a, b]$, and $[b, 1]$:

1. On $[0, a]$: there exists $c_1 \in (0, a)$ such that $g'(c_1) = 0$.
2. On $[a, b]$: there exists $c_2 \in (a, b)$ such that $g'(c_2) = 0$.
3. On $[b, 1]$: there exists $c_3 \in (b, 1)$ such that $g'(c_3) = 0$.

Thus, we have at least three distinct points $c_1, c_2, c_3$ in $(0,1)$ where $g'(x) = 0$. Since $g'(x) = f'(x) - 2$, this means $f'(c_1) = 2$, $f'(c_2) = 2$, and $f'(c_3) = 2$.

Step 8: Apply Rolle's Theorem to $g'(x)$. We have established that $g'(x)$ has at least three distinct roots ($c_1, c_2, c_3$) in the interval $(0,1)$. Since $g'(x)$ is differentiable on $(0,1)$ (because $f''(x)$ exists), we can apply Rolle's Theorem to $g'(x)$ on the intervals $[c_1, c_2]$ and $[c_2, c_3]$.

1. On $[c_1, c_2]$: Since $g'(c_1) = g'(c_2) = 0$, there exists at least one point $d_1 \in (c_1, c_2)$ such that $(g'(x))'(d_1) = 0$. This means $g''(d_1) = 0$. Since $g''(x) = f''(x)$, we have $f''(d_1) = 0$.

2. On $[c_2, c_3]$: Since $g'(c_2) = g'(c_3) = 0$, there exists at least one point $d_2 \in (c_2, c_3)$ such that $(g'(x))'(d_2) = 0$. This means $g''(d_2) = 0$. Since $g''(x) = f''(x)$, we have $f''(d_2) = 0$.

Since $c_1 < c_2 < c_3$, the points $d_1$ and $d_2$ are distinct and lie in $(0,1)$. Therefore, there are at least two distinct points in $(0,1)$ where $f''(x) = 0$.

Step 9: Verify if more points are guaranteed. The problem asks for the least number of points. We have shown that at least 2 points are guaranteed. Could there be more? The given condition is that the line intersects $f(x)$ at only two distinct points in $(0,1)$. This is crucial.

Let's consider the function $h(x) = f(x) - (2x+3)$. We are given that $h(x)$ has exactly two roots in $(0,1)$. Let these roots be $x_1$ and $x_2$, with $0 < x_1 < x_2 < 1$. We also have $h(0) = f(0) - (2(0)+3) = 3 - 3 = 0$. And $h(1) = f(1) - (2(1)+3) = 5 - 5 = 0$. So, $h(x)$ has roots at $0, x_1, x_2, 1$. These are four distinct roots in $[0,1]$.

By Rolle's Theorem applied to $h(x)$ on $[0, x_1]$, $[x_1, x_2]$, and $[x_2, 1]$, there exist $c_1 \in (0, x_1)$, $c_2 \in (x_1, x_2)$, and $c_3 \in (x_2, 1)$ such that $h'(c_1) = 0$, $h'(c_2) = 0$, and $h'(c_3) = 0$. These are three distinct points in $(0,1)$ where $h'(x) = 0$.

Now, apply Rolle's Theorem to $h'(x)$ on the intervals $[c_1, c_2]$ and $[c_2, c_3]$.

1. On $[c_1, c_2]$: Since $h'(c_1) = h'(c_2) = 0$, there exists $d_1 \in (c_1, c_2)$ such that $(h'(x))'(d_1) = 0$, which means $h''(d_1) = 0$.
2. On $[c_2, c_3]$: Since $h'(c_2) = h'(c_3) = 0$, there exists $d_2 \in (c_2, c_3)$ such that $(h'(x))'(d_2) = 0$, which means $h''(d_2) = 0$.

Since $c_1 < c_2 < c_3$, the points $d_1$ and $d_2$ are distinct and lie in $(0,1)$. We have $h''(x) = f''(x)$. Therefore, there are at least two distinct points $d_1, d_2 \in (0,1)$ such that $f''(d_1) = 0$ and $f''(d_2) = 0$.

The least number of points at which $f''(x)=0$ is 2.

Common Mistakes & Tips

• Confusing roots of $f(x)$ with roots of $f'(x)$ or $f''(x)$: Carefully define helper functions and track which function's roots you are analyzing.
• Forgetting the boundary conditions: The values of $f(0)$ and $f(1)$ are crucial for establishing enough roots of the helper function to guarantee the required number of roots for its derivatives.
• Misapplying Rolle's Theorem: Ensure that the conditions for Rolle's Theorem (continuity, differentiability, and equal function values at the endpoints of the interval) are met before applying it.

Summary

We are asked to find the least number of points where $f''(x) = 0$. We define a helper function $g(x) = f(x) - (2x+3)$. The roots of $g(x)$ correspond to the intersection points of $f(x)$ and the line $y=2x+3$. We are given that $f(x)$ intersects $y=2x+3$ at exactly two distinct points in $(0,1)$. Let these points be $x_1$ and $x_2$. Additionally, from the given boundary conditions, $f(0)=3$ and $f(1)=5$, we find that $g(0)=0$ and $g(1)=0$. Thus, $g(x)$ has at least four distinct roots in $[0,1]$: $0, x_1, x_2, 1$. Applying Rolle's Theorem to $g(x)$ on the intervals $[0, x_1]$, $[x_1, x_2]$, and $[x_2, 1]$, we find at least three distinct points where $g'(x)=0$. Applying Rolle's Theorem to $g'(x)$ on the intervals formed by these roots, we find at least two distinct points where $g''(x)=0$. Since $g''(x) = f''(x)$, there are at least two points in $(0,1)$ where $f''(x)=0$.

The final answer is \boxed{2}.