Find the intervals in which the following functions are increasing :
(i) `2x^3-3x`
(ii) `10-6x-2x^(2)`.

Answer

To find the intervals in which the given functions are increasing, we need to determine where the derivative of each function is greater than zero. Let's solve each part step by step.

### Part (i): Function \( f(x) = 2x^3 - 3x \)

**Step 1: Find the derivative of the function.**

\[ f'(x) = \frac{d}{dx}(2x^3 - 3x) = 6x^2 - 3 \]

**Step 2: Set the derivative greater than zero to find intervals of increase.**

\[ 6x^2 - 3 > 0 \]

**Step 3: Solve the inequality.**

First, we can factor the inequality:

\[ 6x^2 > 3 \implies x^2 > \frac{1}{2} \]

Taking the square root of both sides, we get:

\[ |x| > \frac{1}{\sqrt{2}} \implies x < -\frac{1}{\sqrt{2}} \quad \text{or} \quad x > \frac{1}{\sqrt{2}} \]

**Step 4: Identify the critical points.**

The critical points are:

\[ x = -\frac{1}{\sqrt{2}}, \quad x = \frac{1}{\sqrt{2}} \]

**Step 5: Test intervals around the critical points.**

We will test the intervals: \( (-\infty, -\frac{1}{\sqrt{2}}) \), \( (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \), and \( (\frac{1}{\sqrt{2}}, \infty) \).

1. For \( x < -\frac{1}{\sqrt{2}} \) (e.g., \( x = -2 \)):
\[ f'(-2) = 6(-2)^2 - 3 = 24 - 3 = 21 > 0 \]
(Increasing)

2. For \( -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \) (e.g., \( x = 0 \)):
\[ f'(0) = 6(0)^2 - 3 = -3 < 0 \]
(Decreasing)

3. For \( x > \frac{1}{\sqrt{2}} \) (e.g., \( x = 2 \)):
\[ f'(2) = 6(2)^2 - 3 = 24 - 3 = 21 > 0 \]
(Increasing)

**Conclusion for Part (i):**

The function \( f(x) = 2x^3 - 3x \) is increasing on the intervals:

\[ (-\infty, -\frac{1}{\sqrt{2}}) \quad \text{and} \quad (\frac{1}{\sqrt{2}}, \infty) \]

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### Part (ii): Function \( f(x) = 10 - 6x - 2x^2 \)

**Step 1: Find the derivative of the function.**

\[ f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x \]

**Step 2: Set the derivative greater than zero to find intervals of increase.**

\[ -6 - 4x > 0 \]

**Step 3: Solve the inequality.**

Rearranging gives:

\[ -4x > 6 \implies x < -\frac{3}{2} \]

**Conclusion for Part (ii):**

The function \( f(x) = 10 - 6x - 2x^2 \) is increasing on the interval:

\[ (-\infty, -\frac{3}{2}) \]

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### Summary of Intervals of Increase:

1. For \( f(x) = 2x^3 - 3x \): Increasing on \( (-\infty, -\frac{1}{\sqrt{2}}) \) and \( (\frac{1}{\sqrt{2}}, \infty) \).
2. For \( f(x) = 10 - 6x - 2x^2 \): Increasing on \( (-\infty, -\frac{3}{2}) \).

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