To determine the intervals where the bacteria are growing, we need to find where the derivative of the population growth function $f(x) = \ln(-x^2 + 10) + 2(x - 1)$ is positive. A positive derivative indicates that the population is increasing, which corresponds to the bacteria growing.

Step 1: Find the derivative of $f(x)$

We’ll differentiate each part of the function $f(x)$.

1. The derivative of $\ln(-x^2 + 10)$ using the chain rule is: $\frac{d}{dx} \ln(-x^2 + 10) = \frac{1}{-x^2 + 10} \cdot (-2x) = \frac{-2x}{-x^2 + 10}$

2. The derivative of $2(x - 1)$ is simply: $\frac{d}{dx} \left( 2(x - 1) \right) = 2$

Thus, the derivative of $f(x)$ is: $f'(x) = \frac{-2x}{-x^2 + 10} + 2$

Step 2: Determine where $f'(x) > 0$

We need to find where $f'(x)$ is positive. Start by solving the inequality: $\frac{-2x}{-x^2 + 10} + 2 > 0$

Simplify this inequality:

$\frac{-2x}{-x^2 + 10} > -2$

To eliminate the fraction, multiply both sides by $-x^2 + 10$, keeping in mind that it changes the direction of the inequality when $-x^2 + 10$ is negative:

$-2x > -2(-x^2 + 10)$

Simplifying further: $-2x > 2x^2 - 20$ $0 > 2x^2 + 2x - 20$

Divide the inequality by 2 to simplify: $0 > x^2 + x - 10$

Solve the quadratic inequality $x^2 + x - 10 < 0$ by finding the roots of the corresponding quadratic equation: $x^2 + x - 10 = 0$ The solutions to this equation are: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-10)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 40}}{2} = \frac{-1 \pm \sqrt{41}}{2}$ Thus, the roots are approximately: $x \approx \frac{-1 + \sqrt{41}}{2} \approx 3.20 \quad \text{and} \quad x \approx \frac{-1 - \sqrt{41}}{2} \approx -3.20$

Step 3: Analyze the intervals

We now have the quadratic inequality: $x^2 + x - 10 < 0$ This inequality is satisfied between the roots $x \approx -3.20$ and $x \approx 3.20$. Therefore, the population is growing in the interval: $(-\sqrt{41}/2, \sqrt{41}/2)$ which approximates to: $(-3.20, 3.20)$

Conclusion:

The bacterial population is growing on the interval $(-3.20, 3.20)$, assuming that the values of $x$ represent time or some variable where the function is valid.

Would you like a more detailed explanation or have any further questions about this process?

Related Questions:

1. How would the bacterial growth change if the function was $f(x) = \ln(x^2 + 10) + 2(x - 1)$?
2. What does the behavior of the derivative indicate about the rate of bacterial growth at the boundaries of the interval?
3. How does the presence of the logarithmic function affect the growth compared to a simple linear model?
4. Can we interpret the population's growth as exponential in any way from this function?
5. What would happen if we expanded this model to include a carrying capacity for the bacterial population?

Tip:

When analyzing a function’s growth, always pay attention to its derivative. Positive values indicate growth, while negative values indicate decay or decline!