Let λ* be the largest value of λ for which the function f λ(x) =4Ax³ -362x² +36x +48 is increasing for all Then, f*(1) + f*(-1) is equal to

1. Why the derivative decides monotonicity

For a differentiable function $f(x)$:

• If $f'(x)>0$ for every real $x$, the slope is always positive – so the function is strictly increasing.
• If $f'(x)\ge 0$ for every $x$, the function is (non-strictly) increasing.

2. Take the derivative of the given family

The function is
$f_{\lambda}(x)=4\lambda x^{3}-36\lambda x^{2}+36x+48$ Differentiate term-by-term:
$f'_{\lambda}(x)=12\lambda x^{2}-72\lambda x+36$ Factor out the common $12$:
$f'_{\lambda}(x)=12\Big(\lambda x^{2}-6\lambda x+3\Big)$ (The factor 12 is always positive, so the sign of the derivative is controlled by the quadratic inside the brackets.)

3. For which $\lambda$ is the quadratic non-negative for all $x$?

A quadratic $ax^{2}+bx+c$ is $\ge 0$ for every real $x$ when

• $a>0$ (opens upward), and
• its discriminant $\Delta=b^{2}-4ac\le 0$ (so it never crosses the $x$-axis).

Here:
$a=\lambda$, $b=-6\lambda$, $c=3$

1. Upward opening
   $\lambda>0$ (downward opening can’t stay non-negative everywhere unless the parabola is flat).
2. Discriminant condition
   $\Delta = (-6\lambda)^2 - 4(\lambda)(3) = 36\lambda^{2}-12\lambda \le 0$
   Factor out $12\lambda$:
   $12\lambda\,(3\lambda-1)\le 0$
   We already need $\lambda\ge 0$, so the product is non-positive only when
   $3\lambda-1\le 0\quad\Longrightarrow\quad \lambda\le\tfrac13$

Thus the admissible range is
$0\le \lambda\le\tfrac13$ The largest such value is
$\boxed{\lambda^*=\tfrac13}$

4. Evaluate $f_{\lambda^*}(1)+f_{\lambda^*}(-1)$

Substitute $\lambda^*=\tfrac13$ back into $f_{\lambda}(x)$: $f_{1/3}(x)=\frac43x^{3}-12x^{2}+36x+48$ Compute at $x=1$:
$f_{1/3}(1)=\frac43-12+36+48=\frac{4}{3}+72=\frac{220}{3}$

Compute at $x=-1$:
$f_{1/3}(-1)=\frac43(-1)-12(1)-36+48=-\frac43-12-36+48=-\frac43$

Add them:
$f_{1/3}(1)+f_{1/3}(-1)=\frac{220}{3}-\frac{4}{3}=\frac{216}{3}=72$

Hence the required value is
$\boxed{72}$

🧒 Simple way to see the problem

1. We have a family of curves that depend on a number $\lambda$ (say LAMB-da).
2. "Increasing everywhere" means that if you walk along the graph from left to right you never go downhill.
3. To check that, grown-ups look at the derivative (it tells whether the curve is going up or down).
4. We tune $\lambda$ as high as possible but still make sure the derivative is never negative.
5. After finding that best $\lambda$, we just plug x = 1 and x = –1 in the formula and add the two answers together.

The final number turns out to be 72.

Step-by-step calculation

1. Derivative:
   $f'_{\lambda}(x)=12\big(\lambda x^{2}-6\lambda x+3\big)$
2. Quadratic conditions:
   • Upward opening $\Rightarrow \lambda>0$
   • Discriminant $\le 0$:
     $36\lambda^{2}-12\lambda\le0 \;\Rightarrow\; 12\lambda(3\lambda-1)\le0$
     $\Rightarrow\; 0\le\lambda\le\tfrac13$
3. Largest value: $\lambda^{*}=\tfrac13$.
4. Evaluate function:
   $f_{1/3}(x)=\frac43x^{3}-12x^{2}+36x+48$
   $f_{1/3}(1)=\frac{4}{3}-12+36+48=\frac{220}{3}$
   $f_{1/3}(-1)=\frac43(-1)-12-36+48=-\frac43$
   $f_{1/3}(1)+f_{1/3}(-1)=\frac{220}{3}-\frac{4}{3}=\frac{216}{3}=72$

Answer: 72