The question is asking whether the function $f(x) = x^2 + 2x - 3$ is increasing on the interval $(-1, \infty)$ and decreasing on the interval $(-\infty, -1)$.

Step-by-step Analysis:

1. Find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(x^2 + 2x - 3) = 2x + 2$ This derivative tells us the slope of the function at any point.

2. Critical points: To find where the function changes from increasing to decreasing (or vice versa), we set the derivative equal to zero: $f'(x) = 2x + 2 = 0$ Solving for $x$: $2x = -2 \quad \Rightarrow \quad x = -1$ So, $x = -1$ is a critical point.

3. Determine increasing and decreasing intervals:

   • For $x > -1$ (to the right of $-1$): $f'(x) = 2x + 2 > 0 \quad \text{(positive slope)}$ Hence, $f(x)$ is increasing on $(-1, \infty)$.

   • For $x < -1$ (to the left of $-1$): $f'(x) = 2x + 2 < 0 \quad \text{(negative slope)}$ Hence, $f(x)$ is decreasing on $(-\infty, -1)$.

Conclusion:

• The function increases on $(-1, \infty)$ and decreases on $(-\infty, -1)$, so the statement in the question is true. However, the answer marked "False" in the image is incorrect based on this analysis.

Would you like more details or have any questions?

Related questions:

1. What is the significance of critical points in a function's behavior?
2. How do we classify a function's intervals as increasing or decreasing using its derivative?
3. Can a function have more than one critical point?
4. What happens at a point where the derivative is zero but does not change sign?
5. How do second derivatives help in determining concavity?

Tip: Always check the derivative to find the intervals of increase or decrease for any given function.