To find the local maximum of the function $f(x) = x^2 + 3x - 1$ on the interval $[-1, 1]$, we'll follow these steps:

1. Find the derivative of $f(x)$

The first derivative of $f(x)$ is: $f'(x) = \frac{d}{dx} \left(x^2 + 3x - 1\right) = 2x + 3$

2. Set the derivative equal to zero and solve for $x$

To find the critical points: $f'(x) = 2x + 3 = 0$ $2x = -3$ $x = -\frac{3}{2}$

3. Check if the critical point is within the interval $[-1, 1]$

The critical point $x = -\frac{3}{2}$ is not within the interval $[-1, 1]$, so we don't consider it.

4. Evaluate $f(x)$ at the endpoints of the interval

Since there are no critical points within the interval, the local maximum will occur at one of the endpoints:

• At $x = -1$: $f(-1) = (-1)^2 + 3(-1) - 1 = 1 - 3 - 1 = -3$

• At $x = 1$: $f(1) = (1)^2 + 3(1) - 1 = 1 + 3 - 1 = 3$

5. Determine the local maximum

By comparing the values:

• $f(-1) = -3$
• $f(1) = 3$

The local maximum value of $f(x)$ on the interval $[-1, 1]$ is 3, which occurs at $x = 1$.

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Would you like more details or have any questions?

Here are 5 related questions:

1. How do you find the local minimum of a function?
2. What is the difference between a local and a global maximum?
3. How do you find the second derivative to check concavity?
4. Can you explain how to solve optimization problems with constraints?
5. What role does the derivative play in determining critical points?

Tip: Always check the endpoints of an interval when finding local maxima or minima on a closed interval.