Animated Solution for Mathematics - Differentiation: Let $g(x)=3f\left(\frac{x}{3}\right)+f(3-x)$ and $f^{\prime \prime }(x)>0$ for all $x\in (0,3)$. If $g$ is decreasing in $(0,\alpha )$ and increasing in $(\alpha ,3)$, then $8\alpha$ is

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are peeling back the layers of a function to understand its very soul.

We are given $g(x)=3f(\frac{x}{3})+f(3-x)$ and the crucial information that $f^{\prime \prime }(x)>0$ for all $x\in (0,3)$. This condition is the heartbeat of the problem.

The condition $f^{\prime \prime }(x)>0$ tells us that the function $f(x)$ is strictly convex, or "concave up." Imagine a bowl-shaped curve where the slope of the tangent line is constantly increasing.

Mathematically, this implies that the derivative $f^{\prime }(x)$ is a strictly increasing function. For any two points $x_1$ and $x_2$ where $x_1<x_2$, we are guaranteed that $f^{\prime }(x_1)<f^{\prime }(x_2)$. This property is the key that will unlock the entire puzzle.

A Dance of Derivatives

To determine the monotonicity of $g(x)$, we must examine its derivative, $g^{\prime }(x)$. Applying the chain rule with precision, we differentiate the expression term by term.

For the first term, $3f(\frac{x}{3})$, the derivative is $3\cdot f^{\prime }(\frac{x}{3})\cdot \frac{1}{3}$, which simplifies to $f^{\prime }(\frac{x}{3})$. For the second term, $f(3-x)$, the derivative is $f^{\prime }(3-x)\cdot (-1)$.

Combining these, we obtain the following expression:

The function $g(x)$ is decreasing when its derivative is negative. We set $g^{\prime }(x)<0$ and solve the resulting inequality:

Since $f^{\prime }(x)$ is a strictly increasing function, the inequality $f^{\prime }(A)<f^{\prime }(B)$ implies $A<B$. We can therefore "strip away" the $f^{\prime }$ notation and focus on the inputs:

We now solve for $x$ using basic algebraic manipulation:

We have determined that $g(x)$ is decreasing for $x<\frac{9}{4}$. Given the interval $(0,\alpha )$, we identify $\alpha =\frac{9}{4}$.

The final step is to calculate the value of $8\alpha$:

We have successfully navigated the landscape of calculus to reach the result of 18. Always look for the underlying properties of functions to simplify complex problems.