Find the interval where the function f(x) `10 -x^3+6x^2-9x` increases with x.

Options

A. [1,3]
B. (1,3)
C. (1,3]
D. [1,3)

Answer

To find the interval where the function \( f(x) = 10 - x^3 + 6x^2 - 9x \) is increasing, we need to follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(10 - x^3 + 6x^2 - 9x) \]

The derivative of each term is:
- The derivative of \( 10 \) is \( 0 \).
- The derivative of \( -x^3 \) is \( -3x^2 \).
- The derivative of \( 6x^2 \) is \( 12x \).
- The derivative of \( -9x \) is \( -9 \).

Putting it all together, we have:

\[ f'(x) = -3x^2 + 12x - 9 \]

### Step 2: Set the derivative greater than zero
To find where the function is increasing, we set the derivative greater than zero:

\[ -3x^2 + 12x - 9 > 0 \]

### Step 3: Factor the derivative
We can factor out \( -3 \) from the inequality:

\[ -3(x^2 - 4x + 3) > 0 \]

Dividing both sides by \( -3 \) (which reverses the inequality):

\[ x^2 - 4x + 3 < 0 \]

### Step 4: Factor the quadratic expression
Next, we factor the quadratic expression:

\[ x^2 - 4x + 3 = (x - 1)(x - 3) \]

Thus, we have:

\[ (x - 1)(x - 3) < 0 \]

### Step 5: Determine the intervals
To find the intervals where this inequality holds, we can test the sign of the product \( (x - 1)(x - 3) \) in the intervals determined by the roots \( x = 1 \) and \( x = 3 \).

- For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 1)(0 - 3) = 1 \cdot 3 > 0 \)
- For \( 1 < x < 3 \): Choose \( x = 2 \) → \( (2 - 1)(2 - 3) = 1 \cdot (-1) < 0 \)
- For \( x > 3 \): Choose \( x = 4 \) → \( (4 - 1)(4 - 3) = 3 \cdot 1 > 0 \)

### Step 6: Conclusion
The function \( f(x) \) is increasing where \( (x - 1)(x - 3) < 0 \), which occurs in the interval:

\[ (1, 3) \]

### Final Answer:
The function \( f(x) \) is increasing on the interval \( (1, 3) \).

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