To show that the function $f(x) = x^2 + 2x + 4$, defined on the interval $[-1, 2]$, is increasing, we need to analyze its derivative. A function is increasing if its derivative is non-negative (i.e., $f'(x) \geq 0$) for all $x$ in the given interval.

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

The derivative of $f(x) = x^2 + 2x + 4$ with respect to $x$ is:

$f'(x) = \frac{d}{dx}(x^2 + 2x + 4) = 2x + 2$

Step 2: Analyze the sign of $f'(x)$

The derivative $f'(x) = 2x + 2$ is a linear function. To check whether it is non-negative on the interval $[-1, 2]$, let's evaluate its behavior at the endpoints and within the interval:

• At $x = -1$: $f'(-1) = 2(-1) + 2 = 0$

• At $x = 2$: $f'(2) = 2(2) + 2 = 6$

Thus, the derivative is 0 at $x = -1$ and 6 at $x = 2$.

Step 3: General behavior of $f'(x)$

Since $f'(x) = 2x + 2$ is a linear function with a positive slope (2), it increases as $x$ increases. More importantly, for any $x$ in the interval $[-1, 2]$, the derivative $f'(x) \geq 0$.

In particular:

• For $x = -1$, $f'(x) = 0$, which indicates a horizontal tangent (the function is neither increasing nor decreasing at that point).
• For $x > -1$, $f'(x) > 0$, indicating that the function is strictly increasing.

Conclusion:

Since $f'(x) \geq 0$ for all $x \in [-1, 2]$, we can conclude that $f(x)$ is an increasing function on the interval $[-1, 2]$.

Would you like more details on how the derivative affects function behavior?

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Related Questions:

1. What does it mean for a function to be strictly increasing versus increasing?
2. Can a function be increasing if its derivative is 0 at some point?
3. How does concavity affect the increasing or decreasing nature of a function?
4. How can the second derivative provide additional information about function behavior?
5. What is the relationship between the derivative and the graph of a function?

Tip: Always check the sign of the derivative to determine where a function is increasing or decreasing.