Molly's conjecture is that for any number $x > 0$, the inequality $x^3 > 3x + 3$ holds. Let's analyze whether this conjecture is true for all positive $x$.

Step 1: Set up the inequality

$x^3 > 3x + 3$

Step 2: Subtract $3x + 3$ from both sides to bring everything to one side:

$x^3 - 3x - 3 > 0$

Step 3: Analyze the function

Define the function $f(x) = x^3 - 3x - 3$. We want to find out where $f(x) > 0$.

Step 4: Determine the critical points

To determine the behavior of $f(x)$, we first find its derivative and solve for $f'(x) = 0$:

$f'(x) = 3x^2 - 3$

Set $f'(x) = 0$:

$3x^2 - 3 = 0$ $x^2 = 1$ $x = \pm 1$

Since we're only interested in $x > 0$, we consider $x = 1$.

Step 5: Analyze the sign of $f(x)$

Let's check the sign of $f(x)$ in intervals around the critical point $x = 1$.

1. For $0 < x < 1$, pick a test point, say $x = 0.5$: $f(0.5) = (0.5)^3 - 3(0.5) - 3 = 0.125 - 1.5 - 3 = -4.375$ $f(x) < 0$ in this interval.

2. For $x = 1$: $f(1) = 1^3 - 3(1) - 3 = 1 - 3 - 3 = -5$ $f(1) < 0$.

3. For $x > 1$, pick a test point, say $x = 2$: $f(2) = 2^3 - 3(2) - 3 = 8 - 6 - 3 = -1$ $f(x)$ starts to approach zero as $x$ increases but remains negative.

Conclusion:

From this analysis, we see that $f(x) = x^3 - 3x - 3$ is negative for $x > 0$ and doesn't become positive as $x$ increases. Therefore, Molly's conjecture that $x^3 > 3x + 3$ for all $x > 0$ is false.

Would you like to explore this further, or do you have any other questions?

Related Questions:

1. What is the minimum value of the function $f(x) = x^3 - 3x - 3$ for $x > 0$?
2. How does $x^3$ compare to other linear functions like $2x + 1$?
3. Can you find a specific value of $x > 0$ where $x^3 = 3x + 3$?
4. What happens to the inequality $x^3 > 3x + 3$ if $x < 0$?
5. How can we generalize the inequality $x^3 > ax + b$ for other constants $a$ and $b$?

Tip: When dealing with inequalities involving polynomials, finding the critical points and testing intervals is a powerful method for determining where the inequality holds.