The question asks to find the maximum points $X_{\text{max}}$ for the given function:

$y = -\frac{1}{4}(x^3 - 3x + 4)$

Steps to solve:

1. Find the derivative: To find the critical points, we calculate the derivative of $y$: $y' = -\frac{1}{4}(3x^2 - 3)$ Simplify: $y' = -\frac{3}{4}(x^2 - 1)$

2. Find critical points: Set $y' = 0$: $-\frac{3}{4}(x^2 - 1) = 0$ Solve: $x^2 - 1 = 0 \implies x = \pm 1$

3. Determine the nature of critical points: Use the second derivative test or the behavior of $y'$ around the critical points: $y'' = -\frac{3}{4}(2x) = -\frac{3}{2}x$

   • At $x = 1$: $y'' = -\frac{3}{2}(1) = -\frac{3}{2} < 0$, so $x = 1$ is a local maximum.
   • At $x = -1$: $y'' = -\frac{3}{2}(-1) = \frac{3}{2} > 0$, so $x = -1$ is a local minimum.

4. Conclusion: The maximum occurs at $x = 1$. Therefore, the answer is:

   $X_{\text{max}} = 1$

Let me know if you'd like a more detailed explanation or additional assistance!

Related Questions:

1. What is the value of the function at $x = 1$ (maximum point)?
2. How do we classify points using the first derivative test?
3. What happens to the function as $x \to \pm \infty$?
4. How would you find the minimum point for the given function?
5. Could this function have any inflection points? If so, how would you find them?

Tip:

For cubic functions, critical points can always be classified by testing the sign of the derivative on intervals divided by these points.