Let's solve this problem step by step for each part.

Given Function:

$f(x) = x^2$

a) For $-1 < x < 2$ (open interval)

• We first find the critical points by differentiating the function: $f'(x) = 2x$
• Set the derivative equal to 0 to find critical points: $2x = 0 \quad \Rightarrow \quad x = 0$
• Now, check the behavior of $f(x)$ on the interval $-1 < x < 2$. The critical point $x = 0$ lies within this interval.

We evaluate $f(x) = x^2$ at:

• $x = 0$: $f(0) = 0^2 = 0$

• As $x \to -1$ and $x \to 2$, the function approaches: $f(-1) = (-1)^2 = 1 \quad \text{and} \quad f(2) = 2^2 = 4$ But since the interval is open at both ends, we do not consider these boundary points.

• The minimum value in this open interval occurs at $x = 0$ (since $f(0) = 0$), and the maximum value occurs as $x$ approaches 2, with $f(x)$ increasing without bound.

Thus, for $-1 < x < 2$:

• Minimum Value: $0$ at $x = 0$
• Maximum Value: There is no maximum value since $f(x) \to \infty$ as $x \to 2$.

b) For $-1 \leq x \leq 2$ (closed interval)

Here, we consider the boundary points as well.

• Evaluate the function at the boundaries: $f(-1) = (-1)^2 = 1 \quad \text{and} \quad f(2) = 2^2 = 4$

• We already know that the critical point inside the interval is $x = 0$, with $f(0) = 0$.

• Therefore, the values are:

  • At $x = -1$, $f(x) = 1$
  • At $x = 0$, $f(x) = 0$
  • At $x = 2$, $f(x) = 4$

• The minimum value is $0$ at $x = 0$, and the maximum value is $4$ at $x = 2$.

Thus, for $-1 \leq x \leq 2$:

• Minimum Value: $0$ at $x = 0$
• Maximum Value: $4$ at $x = 2$

c) For $0 < x < 2$ (open interval)

• In this case, the critical point $x = 0$ is outside the interval, so we don't need to consider it.

• The function is increasing in this interval because $f'(x) = 2x > 0$ for all $x > 0$.

• We evaluate the function at the boundary points:

  • As $x \to 0^+$, $f(x) \to 0$
  • As $x \to 2$, $f(x) \to 4$

Thus, for $0 < x < 2$:

• Minimum Value: $0$ (approaching as $x \to 0^+$)
• Maximum Value: $4$ as $x \to 2$

Summary of Results:

• a) For $-1 < x < 2$, the minimum value is $0$ at $x = 0$, and there is no maximum value (since $f(x) \to \infty$ as $x \to 2$).
• b) For $-1 \leq x \leq 2$, the minimum value is $0$ at $x = 0$, and the maximum value is $4$ at $x = 2$.
• c) For $0 < x < 2$, the minimum value is $0$ (approaching as $x \to 0^+$), and the maximum value is $4$ at $x = 2$.

Let me know if you'd like more details or clarifications!

Related Questions:

1. How do we find the maximum and minimum values for more complex functions, like polynomials of higher degrees?
2. What is the significance of the first derivative in finding critical points?
3. How do we handle cases where the function is not continuous on the interval?
4. What are the differences between open and closed intervals in terms of maximum and minimum values?
5. Can we apply these same methods to trigonometric or logarithmic functions?

Tip:

When finding maximum and minimum values, always check both the critical points (where the derivative equals zero) and the boundary points of the interval, especially in closed intervals!