Let λ* be the largest value of λ for which the function f λ(x) =4λx³ -36λx² +36x +48 is increasing for all Then, f*(1) + f*(-1) is equal to

1. Derivative gives the slope

For a function $f(x)$ to be increasing on the entire real line $\mathbb R$, its derivative must satisfy

$f'(x) \ge 0 \quad \text{for all } x \in \mathbb R.$

Given

$f_\lambda(x)=4\lambda x^3-36\lambda x^2+36x+48,$

we differentiate term-by-term:

$f_\lambda'(x)=\frac{d}{dx}\left(4\lambda x^3\right)-\frac{d}{dx}\left(36\lambda x^2\right)+\frac{d}{dx}(36x)+\frac{d}{dx}(48).$

Carrying it out:

$f_\lambda'(x)=12\lambda x^2-72\lambda x+36.$

To simplify analysis pull out a common factor of 12:

$f_\lambda'(x)=12\bigl(\lambda x^2-6\lambda x+3\bigr).$

2. Force the quadratic to be non-negative everywhere

Because $12>0$, the sign of $f_\lambda'(x)$ is entirely controlled by the quadratic

$g(x)=\lambda x^2-6\lambda x+3.$

A quadratic $ax^2+bx+c$ is non-negative for all real $x$ when:

1. $a>0$ (opens upward) and
2. its discriminant satisfies $\Delta=b^2-4ac\le 0.$

For $g(x)$:

• $a=\lambda$ ⟹ we need $\lambda>0$.
• $b=-6\lambda, \; c=3$.

Compute the discriminant:

$\Delta=(-6\lambda)^2-4(\lambda)(3)=36\lambda^2-12\lambda=12\lambda\,(3\lambda-1).$

Set $\Delta\le 0$:

$12\lambda\,(3\lambda-1)\le 0.$

Because $12>0$, divide by $12$ without changing the sign:

$\lambda\,(3\lambda-1)\le 0.$

With $\lambda>0$, the product is (\le 0) only when

$3\lambda-1\le 0 \quad\Longrightarrow\quad \lambda\le\frac{1}{3}.$

Hence the allowed interval is

$0<\lambda\le \frac{1}{3}.$

The largest such value is

$\boxed{\lambda^*=\dfrac{1}{3}}.$

3. Evaluate $f_{\lambda^*}(1)$ and $f_{\lambda^*}(-1)$

Substitute $\lambda^*=\frac{1}{3}$ into the original function:

🧒🏼 What’s the problem saying?

We have a family of curves (think of many slides in a slide-show). Each slide is made by putting a number λ (lambda) into the formula

$f_\lambda(x)=4\lambda x^3-36\lambda x^2+36x+48.$

The question is:

1. Pick the biggest λ for which the curve keeps climbing everywhere (never goes down).
2. Take that special λ, plug it back into the formula, find the value at $x = 1$ and at $x = -1$, then add them.

🪜 How would a kid think of “always climbing”?

Imagine riding a bike on a road: if the road is always going upward, you never go downhill. Mathematicians call that the function being increasing everywhere. To check if a road always goes up we look at its slope (the derivative). If the slope is never negative, the road never goes down!

So we:

1. Find the slope.
2. Force it to stay positive for every spot on the road (every real $x$).
3. That gives the biggest safe λ.
4. Finally do two easy plugs and one addition.

That’s the whole story! 🚲⬆️

Step-by-Step Solution

1. Differentiate $f_\lambda(x)=4\lambda x^3-36\lambda x^2+36x+48$ $f_\lambda'(x)=12\lambda x^2-72\lambda x+36=12\bigl(\lambda x^2-6\lambda x+3\bigr).$

2. Require monotonic increase: $f_\lambda'(x)\ge0 \;\forall x \in \mathbb R\;\Longrightarrow\; \lambda x^2-6\lambda x+3\ge0 \;\forall x.$

3. Quadratic condition (opens upward and non-positive discriminant):

   • $\lambda>0$
   • $\Delta=36\lambda^2-12\lambda\le0 \implies 12\lambda(3\lambda-1)\le0 \implies 0<\lambda\le\dfrac13$.

4. Largest acceptable value: $\lambda^*=\frac13.$

5. Evaluate at $x=1$ and $x=-1$ $f_{\frac13}(x)=\frac{4}{3}x^3-12x^2+36x+48.$ $f_{\frac13}(1)=\frac{4}{3}-12+36+48=\frac{220}{3},$ $f_{\frac13}(-1)=-\frac{4}{3}-12-36+48=-\frac{4}{3}.$

6. Add the two values $f_{\frac13}(1)+f_{\frac13}(-1)=\frac{220}{3}-\frac{4}{3}=\frac{216}{3}=72.$

[ \boxed{72} ]