Animated Solution for Mathematics - Differentiation: Let ${\lambda }^{\ast }$ be the largest value of $\lambda$ for which the function $f_{\lambda }(x)=4\lambda x^3-36\lambda x^2+36x+48$ is increasing for all $x\in \mathbb{R}$. Then $f_{{\lambda }^{\ast }}(1)+f_{{\lambda }^{\ast }}(-1)$ is equal to :

Understanding Monotonicity

Imagine you are standing at the base of a mountain range. You are tasked with walking along a path defined by the function $f_{\lambda }(x)=4\lambda x^3-36\lambda x^2+36x+48$.

The condition is simple but strict: you must never walk downhill. Every step you take must be either level or uphill.

In the language of calculus, this means the slope of your path—the derivative—must never be negative. It must be greater than or equal to zero for every single point $x$ on the real number line.

This is the essence of a monotonically increasing function. It is not just about the function values; it is about the rate of change. Let us begin our journey by finding that rate of change.

The Calculus Bridge

To find the slope, we apply the power rule to our cubic polynomial. Differentiating $f_{\lambda }(x)$ with respect to $x$, we get:

Applying the rule term by term, the $4\lambda x^3$ becomes $12\lambda x^2$, the $-36\lambda x^2$ becomes $-72\lambda x$, and the $36x$ becomes $36$. The constant $48$ vanishes, as it contributes nothing to the slope.

We are left with:

To make our analysis cleaner, let us factor out the common $12$:

We need this expression to be non-negative for all $x\in \mathbb{R}$. This is where the physics of the graph takes over.

The Quadratic Trap

We are now looking at a quadratic expression: $\lambda x^2-6\lambda x+3$. For this to be non-negative for all $x$, the parabola must open upwards, and it must not cross the $x$-axis.

First, for the parabola to open upwards, the leading coefficient must be positive: $\lambda >0$.

Second, to ensure it never dips below the $x$-axis, the discriminant $D$ must be less than or equal to zero. If $D$ were positive, the parabola would cross the axis, creating a region where the slope is negative—and we know that is forbidden on our mountain climb!

So, we set $D=b^2-4ac\le 0$:

Expanding this, we get:

Factoring this inequality, we find:

Since we already established that $\lambda >0$, we can divide by $12\lambda$ without flipping the inequality sign. This leaves us with $3\lambda -1\le 0$, or $\lambda \le \frac{1}{3}$.

Combining this with our condition $\lambda >0$, we find the valid range for $\lambda$ is $(0,\frac{1}{3}]$. The largest value, our ${\lambda }^{\ast }$, is exactly $\frac{1}{3}$.

The Final Evaluation

Now that we have unlocked the value of ${\lambda }^{\ast }=\frac{1}{3}$, we substitute it back into our original function to see the path clearly:

The problem asks us to calculate $f_{{\lambda }^{\ast }}(1)+f_{{\lambda }^{\ast }}(-1)$. Let us compute these values carefully.

For $x=1$:

For $x=-1$:

When we sum these two results, the beauty of the problem reveals itself. The fractional terms $\frac{4}{3}$ and $-\frac{4}{3}$ cancel out perfectly, leaving us with:

And there we have it. Through the rigorous application of calculus and the geometric intuition of parabolas, we have navigated the path and arrived at the summit.

The final answer is 72.