Animated Solution for Mathematics - Differentiation: Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty ,\infty )$ such that $f(x)=f(1-x)$ and $f^{\prime }(\frac{1}{4})=0$. Then,

Imagine you are standing before a perfectly polished mirror. You raise your left hand, and your reflection raises its right. This is the essence of symmetry.

In the world of calculus, when we are given a function satisfying $f(x)=f(1-x)$, we are essentially looking at a mathematical mirror. The function is perfectly symmetric about the vertical line $x=\frac{1}{2}$.

This is not just a geometric curiosity; it is a powerful constraint that dictates the behavior of the function's derivatives and integrals. Let us embark on a journey to decode this symmetry.

The Slope at the Center

Our first task is to understand how the slope of this function behaves. We start with our symmetry equation:

To find the slope, we differentiate both sides with respect to $x$. The left side is straightforward: $\frac{d}{dx}[f(x)]=f^{\prime }(x)$.

On the right side, we must invoke the chain rule. The derivative of $f(1-x)$ is $f^{\prime }(1-x)$ multiplied by the derivative of the inner function $(1-x)$, which is $-1$. Thus, we arrive at the elegant relationship:

This tells us that the slope at any point $x$ is the negative of the slope at its mirror image $1-x$. Now, what happens at the mirror itself, $x=\frac{1}{2}$?

Substituting $x=\frac{1}{2}$ into our derivative equation, we get $f^{\prime }(\frac{1}{2})=-f^{\prime }(1-\frac{1}{2})$, which simplifies to $f^{\prime }(\frac{1}{2})=-f^{\prime }(\frac{1}{2})$. Adding $f^{\prime }(\frac{1}{2})$ to both sides, we find $2f^{\prime }(\frac{1}{2})=0$, which means $f^{\prime }(\frac{1}{2})=0$.

The tangent line at the center of symmetry is perfectly horizontal. This confirms that Option (b) is correct.

The problem provides us with another vital piece of information: $f^{\prime }(\frac{1}{4})=0$. Using our symmetry relation $f^{\prime }(x)=-f^{\prime }(1-x)$, let us test the point $x=\frac{3}{4}$.

We find $f^{\prime }(\frac{3}{4})=-f^{\prime }(1-\frac{3}{4})=-f^{\prime }(\frac{1}{4})$. Since $f^{\prime }(\frac{1}{4})=0$, it follows that $f^{\prime }(\frac{3}{4})=0$.

We now have three distinct points where the first derivative vanishes: $x=\frac{1}{4}$, $x=\frac{1}{2}$, and $x=\frac{3}{4}$. This is where Rolle's Theorem becomes our most trusted ally.

Rolle's Theorem guarantees that between any two roots of a differentiable function, there exists at least one root of its derivative. Applying this to $f^{\prime }(x)$ on the interval $[\frac{1}{4},\frac{1}{2}]$, there must exist some $c_1$ such that $f^{\prime \prime }(c_1)=0$.

Similarly, on the interval $[\frac{1}{2},\frac{3}{4}]$, there must exist some $c_2$ such that $f^{\prime \prime }(c_2)=0$. Thus, $f^{\prime \prime }(x)$ vanishes at least twice on the interval $[0,1]$. Option (a) is confirmed.

Finally, let us look at the integrals. For Option (c), we consider:

By substituting $t=x+\frac{1}{2}$, the limits transform from $[-\frac{1}{2},\frac{1}{2}]$ to $[0,1]$. The integral becomes:

Here, $f(t)$ is symmetric about $t=\frac{1}{2}$, while $\sin (t-\frac{1}{2})$ is anti-symmetric (odd) about $t=\frac{1}{2}$. The product of an even function and an odd function is odd, and the integral of an odd function over a symmetric interval is zero.

Thus, Option (c) is correct. For Option (d), a simple substitution $u=1-t$ reveals that the two sides of the equation are identical.

We have traversed the landscape of this function, from its mirror-like symmetry to the vanishing slopes and the elegant cancellation of integrals. Every step was guided by the fundamental properties of symmetry.