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Prompt: Consider the equation $x^2 + \frac{9}{x^2} = 10$. Solve for $x$ such that the inequality $x^3 - 3x > 0$ is also satisfied. Additionally, show that $y = x^3 - 3x$ is increasing at $x = 1$.

Now, we plug in $x = 1$ into the inequality: $1^3 - 3(1) = 1 - 3 = -2 < 0$. So, $x = 1$ does not satisfy the inequality.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Finally, we plug in $x = -1$ into the inequality: $(-1)^3 - 3(-1) = -1 + 3 = 2 > 0$. So, $x = -1$ satisfies the inequality.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Therefore, the solutions to the equation $x^2 + \frac{9}{x^2} = 10$ that also satisfy the inequality $x^3 - 3x > 0$ are $x = 3$ and $x = -1$.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Now, we need to show that $y = x^3 - 3x$ is increasing at $x = 1$. To do this, we can find the derivative of $y$ with respect to $x$ and evaluate it at $x = 1$.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

The derivative of $y$ with respect to $x$ is given by $\frac{dy}{dx} = 3x^2 - 3$. Evaluating this at $x = 1$, we get $\frac{dy}{dx} = 3(1)^2 - 3 = 3 - 3 = 0$.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

However, we need to check if the derivative is positive or negative in the neighborhood of $x = 1$. Let's evaluate the derivative at $x = 1 + \epsilon$ and $x = 1 - \epsilon$, where $\epsilon$ is a small positive number.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Evaluating the derivative at $x = 1 + \epsilon$, we get $\frac{dy}{dx} = 3(1 + \epsilon)^2 - 3 = 3(1 + 2\epsilon + \epsilon^2) - 3 = 6\epsilon + 3\epsilon^2$. Since $\epsilon$ is positive, this expression is positive.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Evaluating the derivative at $x = 1 - \epsilon$, we get $\frac{dy}{dx} = 3(1 - \epsilon)^2 - 3 = 3(1 - 2\epsilon + \epsilon^2) - 3 = -6\epsilon + 3\epsilon^2$. Since $\epsilon$ is positive, this expression is negative.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Therefore, the derivative of $y$ with respect to $x$ is increasing at $x = 1$, since it changes from negative to positive as $x$ increases through $1$.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate