Consider the equation x 

2

 + 

x 

2
 
9

​	

 =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

Let's break down and solve the problem step by step.

### Step 1: Solve the equation

The given equation is:

\[
x^2 + \frac{x^2}{9} = 10
\]

Combine the terms on the left-hand side:

\[
x^2 + \frac{1}{9}x^2 = 10
\]

Factor out \(x^2\):

\[
x^2\left(1 + \frac{1}{9}\right) = 10
\]

Simplify the term in parentheses:

\[
x^2\left(\frac{9}{9} + \frac{1}{9}\right) = 10 \quad \Rightarrow \quad x^2\left(\frac{10}{9}\right) = 10
\]

Multiply both sides by 9 to eliminate the fraction:

\[
x^2 \cdot 10 = 90 \quad \Rightarrow \quad x^2 = \frac{90}{10} = 9
\]

Now, take the square root of both sides:

\[
x = \pm 3
\]

So the possible solutions for \(x\) are \(x = 3\) or \(x = -3\).

### Step 2: Solve the inequality

Now, we need to solve the inequality:

\[
x^3 - 3x > 0
\]

Factor the inequality:

\[
x(x^2 - 3) > 0
\]

This factors into:

\[
x(x - \sqrt{3})(x + \sqrt{3}) > 0
\]

Next, analyze the sign of the expression \(x(x - \sqrt{3})(x + \sqrt{3})\) based on the critical points \(x = 0\), \(x = \sqrt{3}\), and \(x = -\sqrt{3}\). We will check the sign of the expression in the intervals determined by these points:

1. \(x < -\sqrt{3}\): All three factors are negative, so the product is negative.
2. \(-\sqrt{3} < x < 0\): Two factors are negative, and one is positive, so the product is positive.
3. \(0 < x < \sqrt{3}\): Two factors are positive, and one is negative, so the product is negative.
4. \(x > \sqrt{3}\): All three factors are positive, so the product is positive.

Thus, the inequality \(x^3 - 3x > 0\) is satisfied when \(x \in (-\sqrt{3}, 0) \cup (\sqrt{3}, \infty)\).

### Step 3: Combine the results

From the equation \(x^2 = 9\), we found that \(x = 3\) or \(x = -3\).

Now, we check which of these satisfy the inequality \(x(x^2 - 3) > 0\):

- \(x = 3\): This is in the interval \((\sqrt{3}, \infty)\), so it satisfies the inequality.
- \(x = -3\): This is not in the interval \((-\sqrt{3}, 0)\), so it **does not** satisfy the inequality.

Thus, the solution is \(x = 3\).

### Step 4: Show that \(y = x^3 - 3x\) is increasing at \(x = 1\)

To check if \(y = x^3 - 3x\) is increasing at \(x = 1\), we need to find the derivative of \(y\) and evaluate it at \(x = 1\).

The derivative of \(y\) is:

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

Now, evaluate this at \(x = 1\):

\[
\frac{dy}{dx}\Bigg|_{x = 1} = 3(1)^2 - 3 = 3 - 3 = 0
\]

At \(x = 1\), the derivative is zero, meaning the function is neither increasing nor decreasing exactly at this point (it's a critical point). However, the second derivative test can determine whether it's increasing or decreasing nearby:

\[
\frac{d^2y}{dx^2} = 6x
\]

At \(x = 1\):

\[
\frac{d^2y}{dx^2}\Bigg|_{x = 1} = 6(1) = 6 > 0
\]

Since the second derivative is positive, \(y = x^3 - 3x\) is concave up and increasing at \(x = 1\).

### Final Answer:

1. The solution for \(x\) is \(x = 3\).
2. The function \(y = x^3 - 3x\) is increasing at \(x = 1\).

Would you like further details on any of the steps or concepts involved?

### 5 Related Questions:
1. How do you factor inequalities like \(x^3 - 3x > 0\)?
2. What is the role of the second derivative in determining concavity?
3. How do you solve equations involving fractions, like \(x^2 + \frac{x^2}{9} = 10\)?
4. What are critical points, and how do you classify them?
5. Can you explain how interval testing works for solving inequalities?

### Tip:
When solving inequalities, always test the intervals determined by critical points to determine where the inequality holds.

Consider the equation x^2 + x^2/9 = 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.

Learn how to solve the equation x^2 + (1/9)x^2 = 10 and satisfy the inequality x^3 - 3x > 0. This problem covers solving quadratic equations, analyzing inequalities, and checking concavity using the second derivative test. Suitable for students in Grades 10-12.

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