Let's break this down step-by-step:

1. Solve for $x$ from the equation:

$x^2 + \frac{9}{x^2} = 10$

We multiply both sides by $x^2$ to clear the fraction: $x^4 + 9 = 10x^2$

Rearrange to form a quadratic equation in terms of $u = x^2$: $x^4 - 10x^2 + 9 = 0$

Now, substitute $u = x^2$, so the equation becomes: $u^2 - 10u + 9 = 0$

Solve this quadratic using the quadratic formula: $u = \frac{10 \pm \sqrt{10^2 - 4 \cdot 9}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}$

So: $u = \frac{10 + 8}{2} = 9 \quad \text{and} \quad u = \frac{10 - 8}{2} = 1$

Thus: $x^2 = 9 \quad \text{or} \quad x^2 = 1$

Therefore: $x = \pm 3 \quad \text{or} \quad x = \pm 1$

2. Apply the inequality condition:

We are given that $x^3 - 3x > 0$. Let’s first factor the expression: $x^3 - 3x = x(x - \sqrt{3})(x + \sqrt{3})$

We need the product to be positive. This depends on the sign of each factor across different ranges of $x$. We solve for:

• Sign analysis of $x(x - \sqrt{3})(x + \sqrt{3})$:
  • For $x > \sqrt{3}$: All factors are positive.
  • For $-\sqrt{3} < x < \sqrt{3}$: Signs alternate; the product is negative.
  • For $x < -\sqrt{3}$: All factors are negative, so the product is negative.

Thus, the inequality $x^3 - 3x > 0$ holds only when $x > \sqrt{3}$ or $x < -\sqrt{3}$.

3. Combining the Solutions:

From the equation $x^2 + \frac{9}{x^2} = 10$, the possible solutions are $x = \pm 3$ and $x = \pm 1$. Now, we need to keep only those solutions that satisfy the inequality $x^3 - 3x > 0$.

• For $x = 3$: $3^3 - 3(3) = 27 - 9 = 18 > 0$.
• For $x = -3$: $(-3)^3 - 3(-3) = -27 + 9 = -18 < 0$.
• For $x = \pm 1$: $1^3 - 3(1) = -2 < 0$, $(-1)^3 - 3(-1) = 2 > 0$, but the sign is opposite to the positive requirement.

Thus, the only valid solution is: $x = 3$

4. Show that $y = x^3 - 3x$ is increasing at $x = 1$:

We need to compute the derivative of $y = x^3 - 3x$ to check if it is increasing at $x = 1$.

$\frac{dy}{dx} = 3x^2 - 3$

Evaluate the derivative at $x = 1$: $\frac{dy}{dx} \Big|_{x = 1} = 3(1)^2 - 3 = 3 - 3 = 0$

Since the derivative is 0 at $x = 1$, the function is neither increasing nor decreasing at this point (it is a critical point).

Summary:

• The solution to the equation $x^2 + \frac{9}{x^2} = 10$ that also satisfies the inequality $x^3 - 3x > 0$ is: $x = 3$
• The derivative of $y = x^3 - 3x$ at $x = 1$ is 0, so the function is neither increasing nor decreasing at $x = 1$.

Do you have any questions or need further details?

Related Questions:

1. How do we determine intervals of increase and decrease for a function?
2. Can a function have multiple solutions that partially satisfy a given inequality?
3. What is the significance of critical points in calculus?
4. How does the quadratic formula generalize to higher-degree polynomials?
5. How can we visualize inequalities graphically?

Tip:

When analyzing inequalities involving polynomials, sign charts can be a useful tool to identify solution intervals accurately.