Let's break down the problem in two parts:

Part 1: Solving the equation $x^2 + \frac{9}{x^2} = 10$

We need to solve for $x$ such that the inequality $x^3 - 3x > 0$ is also satisfied.

Step 1: Solve the equation $x^2 + \frac{9}{x^2} = 10$

Multiply both sides by $x^2$ to eliminate the fraction:

$x^4 + 9 = 10x^2$

Rearranging terms gives:

$x^4 - 10x^2 + 9 = 0$

Let $z = x^2$, so the equation becomes a quadratic in $z$:

$z^2 - 10z + 9 = 0$

Solve this quadratic equation using the quadratic formula:

$z = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)}$

$z = \frac{10 \pm \sqrt{100 - 36}}{2}$

$z = \frac{10 \pm \sqrt{64}}{2}$

$z = \frac{10 \pm 8}{2}$

Thus, $z = 9$ or $z = 1$.

Since $z = x^2$, we have two cases:

1. $x^2 = 9$ → $x = \pm 3$
2. $x^2 = 1$ → $x = \pm 1$

So, $x = -3, -1, 1, 3$.

Step 2: Satisfy the inequality $x^3 - 3x > 0$

Now we need to check which of these solutions satisfy $x^3 - 3x > 0$.

Factor the inequality:

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

This implies:

$x(x^2 - 3) > 0$

The solutions are determined by analyzing the sign changes:

• $x(x^2 - 3) = 0$ when $x = 0$ or $x = \pm \sqrt{3}$.
• Test intervals based on these roots: $(-\infty, -\sqrt{3}), (-\sqrt{3}, 0), (0, \sqrt{3}), (\sqrt{3}, \infty)$.

In the interval $(0, \sqrt{3})$, the inequality is negative. In the intervals $(-\infty, -\sqrt{3})$ and $(\sqrt{3}, \infty)$, it is positive.

Now, check which values from our solutions for $x$ satisfy this:

• $x = 3$: $3(3^2 - 3) = 3(9 - 3) = 18 > 0$ (satisfies inequality).
• $x = -3$: $-3((-3)^2 - 3) = -3(9 - 3) = -18 < 0$ (does not satisfy inequality).
• $x = 1$: $1(1^2 - 3) = 1(1 - 3) = -2 < 0$ (does not satisfy inequality).
• $x = -1$: $-1((-1)^2 - 3) = -1(1 - 3) = 2 > 0$ (satisfies inequality).

Thus, $x = 3$ and $x = -1$ satisfy both the equation and the inequality.

Part 2: Show that $y = x^3 - 3x$ is increasing at $x = 1$

To determine whether $y = x^3 - 3x$ is increasing at $x = 1$, we need to check the derivative $y'$.

Step 1: Find the derivative

Differentiate $y = x^3 - 3x$:

$y' = 3x^2 - 3$

Step 2: Evaluate the derivative at $x = 1$

$y'(1) = 3(1)^2 - 3 = 3 - 3 = 0$

Since $y'(1) = 0$, the function is neither increasing nor decreasing at $x = 1$. Thus, at $x = 1$, the function has a critical point, possibly a local minimum or maximum.

Let me know if you need more details or further clarification on any part of the solution!

5 Related Questions:

1. How do you find critical points of a function based on its derivative?
2. Can you graph $y = x^3 - 3x$ and identify where it increases or decreases?
3. What role do the roots of the equation $x^2 + \frac{9}{x^2} = 10$ play in determining the behavior of the function?
4. How do you test intervals to solve inequalities involving cubic functions?
5. Why does the derivative being zero at a point suggest a possible extremum?

Tip:

When solving inequalities, factor the expression to identify critical points and use a sign chart to test different intervals for sign changes.