What are the maximum values of the expression x²⁴ + x⁹ + 2x³ when x is a real number?

To find the maximum values of the expression x²⁴ + x⁹ + 2x³ for real values of x, we need to analyze the behavior of the function.

First, we denote the function as:

f(x) = x²⁴ + x⁹ + 2x³

Next, we can differentiate the function to find its critical points. We will compute the first derivative:

f'(x) = 24x²³ + 9x⁸ + 6x²

Now, we set the first derivative equal to zero to find critical points:

24x²³ + 9x⁸ + 6x² = 0

Factoring out the common term, we get:

6x²(4x² + 3x⁶ + 1) = 0

This gives us two potential solutions:

• 6x² = 0 which gives x = 0
• 4x² + 3x⁶ + 1 = 0 which we will analyze further

The second equation, 4x² + 3x⁶ + 1 = 0, is always positive for real numbers because both 4x² and 3x⁶ are non-negative. Thus, there are no real solutions from this factor.

Next, we evaluate the function at the identified critical point, x = 0:

f(0) = 0²⁴ + 0⁹ + 2(0)³ = 0

To determine if this is a maximum or minimum, let’s analyze the behavior of f(x) as x approaches positive and negative infinity:

• As x → +∞, f(x) → +∞
• As x → -∞, f(x) → +∞ (since the even power dominates)

Since the function approaches positive infinity in both directions and attains a minimum value of 0 at x = 0, it shows that the function does not have a finite maximum value for x in real numbers.

Conclusion: The expression x²⁴ + x⁹ + 2x³ can take any non-negative value but does not have a maximum bound. Its minimum value (which occurs at x = 0) is 0.