Animated Solution for Mathematics - Differentiation: For all $x$ in $[0,1]$, let the second derivative $f^{\prime \prime }(x)$ of a function $f(x)$ exist and satisfy $|f^{\prime \prime }(x)|<1$. If $f(0)=f(1)$, then show that $|f^{\prime }(x)|<1$ for all $x$ in $[0,1]$.

Imagine you are standing on a gentle, rolling hill. You start at point $x=0$ and end at point $x=1$, and the problem states that you are at the exact same elevation at both ends.

This is the physical reality of our function $f(x)$ where $f(0)=f(1)$. This condition serves as the foundation for our exploration.

When a function starts and ends at the same height, it must have turned around somewhere. If you climb a hill and then descend to the same level, there must be a peak or a valley where your path is perfectly flat for an instant.

This is the intuitive heart of Rolle's Theorem. Since our function $f(x)$ is continuous and differentiable, there must exist at least one point $c$ in the interval $(0,1)$ where the slope is zero.

Mathematically, we express this as:

This point $c$ acts as our anchor, providing a reference point amidst the variables.

To understand the slope at any arbitrary point $x$, we connect it to our known slope at $c$ using the Lagrange Mean Value Theorem (LMVT). We apply this theorem not to $f(x)$, but to the derivative function $g(x)=f^{\prime }(x)$.

By applying LMVT to $f^{\prime }(x)$ on the interval between $c$ and $x$, we analyze the rate of change of the slope itself. The theorem guarantees an intermediate point $\xi$ between $c$ and $x$ such that:

This equation is the key to the derivation. Substituting the known value $f^{\prime }(c)=0$ into our LMVT equation yields:

This simplifies beautifully to:

We have now expressed the first derivative $f^{\prime }(x)$ as the product of the second derivative $f^{\prime \prime }(\xi )$ and the distance $(x-c)$.

The problem provides the constraint $|f^{\prime \prime }(x)|<1$ for all $x$. This implies that for any $\xi$, the magnitude $|f^{\prime \prime }(\xi )|$ is strictly less than $1$.

Furthermore, since both $x$ and $c$ reside in the interval $[0,1]$, the distance $|x-c|$ must also be strictly less than $1$. Taking the absolute value of our derived equation, we obtain:

Since both factors on the right side are strictly less than $1$, their product must also be strictly less than $1$. Thus, we conclude:

This result demonstrates how constraints on the curvature of a function dictate the limits of its slope.