Extreme Value Theorem, Extrema, Critical Points - AP Calculus AB

1. State the Second Derivative Test.

   Use $f''(x)$ to determine concavity and local extrema. Evaluate second derivative at critical points.

2. Determine the global extrema of $f(x) = -x^2 + 4x$ on $[0, 4]$.

   Global max at $x = 2$; global min at $x = 0, 4$. Vertex of parabola at $x = 2$ gives maximum.

3. Identify the critical points of $f(x) = \frac{x^3}{3} - x$.

   Critical points at $x = -1, 1$. $f'(x) = x^2 - 1 = 0$ when $x = \pm 1$.

4. Define a critical point of a function.

   A point where $f'(x) = 0$ or $f'(x)$ is undefined. These are the only candidates for local extrema.

5. What does $f''(x) > 0$ at a point indicate?

   The function is concave up; possible local min. Positive second derivative indicates upward concavity.

6. What is a local maximum?

   A point higher than all nearby points. Also called a relative maximum.

7. What is verified by the Second Derivative Test?

   Concavity and nature of local extrema. Second derivative test classifies critical point behavior.

8. What is the impact of a discontinuity on EVT?

   Discontinuity can prevent existence of global extrema. EVT doesn't apply if function isn't continuous.

9. Differentiate between global and local extrema.

   Global is on entire domain; local is on a neighborhood. Global considers entire domain; local considers neighborhoods.

10. What role do endpoints play in finding global extrema?

    Endpoints are evaluated for possible global extrema. Endpoints must be checked for global extrema.

11. What condition must be met for the EVT to apply?

    Function must be continuous on a closed interval. Requires both continuity and a closed, bounded interval.

12. What function property is tested with $f''(x) = 0$?

    Possible inflection point; concavity change. Second derivative zero suggests possible inflection point.

13. What is a global minimum?

    The lowest value of a function on its entire domain. Also called the absolute minimum value.

14. Explain how to find global extrema on a closed interval.

    Evaluate $f(x)$ at critical points and endpoints. Check critical points and interval endpoints systematically.

15. What is a local minimum?

    A point lower than all nearby points. Also called a relative minimum.

16. How can you confirm a global extremum?

    Compare function values at critical points and endpoints. Test all candidates to find absolute extrema.

17. Determine the global extrema of $f(x) = -x^2$ on $[-1, 1]$.

    Global max at $x = 0$; global min at $x = -1, 1$. Parabola opens downward with vertex at origin.

18. What is a sufficient condition for a local minimum?

    $f'(x)$ changes from negative to positive. Derivative changes from negative to positive.

19. Identify the global extrema of $f(x) = x^3 - 3x$ on $[-2, 2]$.

    Global max at $x = 2$; global min at $x = -2$. $f(-2) = -2$, $f(1) = -2$, $f(2) = 2$.

20. Determine the global extrema of $f(x) = x^2 - 4$ on $[-2, 2]$.

    Global max at $x = -2, 2$; global min at $x = 0$. Evaluate at critical point $x = 0$ and endpoints.

21. What is the purpose of the Extreme Value Theorem?

    To ensure global extrema exist on a closed interval. Guarantees extrema exist for optimization problems.

22. What is a global maximum?

    The highest value of a function on its entire domain. Also called the absolute maximum value.

23. Find critical points for $f(x) = \frac{1}{3}x^3 - x^2 + 2$.

    Critical points at $x = 0$ and $x = 2$. $f'(x) = x^2 - 2x = x(x-2) = 0$ when $x = 0, 2$.

24. How do you find critical points?

    Solve $f'(x) = 0$ or find where $f'(x)$ is undefined. Find where the derivative equals zero or doesn't exist.

25. State the First Derivative Test.

    Use $f'(x)$ sign changes to determine local extrema. Check if derivative changes sign around critical points.

26. Why are continuous functions important in EVT?

    Continuity ensures the existence of extrema on $[a, b]$. Discontinuous functions may not have global extrema.

27. What is the significance of critical points?

    Potential locations of local extrema. Only at critical points can local extrema occur.

28. What does $f''(x) < 0$ at a point indicate?

    The function is concave down; possible local max. Negative second derivative indicates downward concavity.

29. What is the Extreme Value Theorem?

    If a function is continuous on [a, b], it attains a global max and min. Guarantees existence of absolute extrema on closed intervals.

30. Identify the critical points of $f(x) = x^2 + 3x + 2$.

    Critical point at $x = -\frac{3}{2}$. $f'(x) = 2x + 3 = 0$ when $x = -\frac{3}{2}$.

31. How does $f'(x)$ determine extrema?

    Sign changes in $f'(x)$ indicate local extrema. Sign changes in first derivative locate extrema.

32. Identify the critical points of $f(x) = x^2 - 4x + 4$.

    Critical point at $x = 2$. $f'(x) = 2x - 4 = 0$ when $x = 2$.

33. Identify the local extrema of $f(x) = x^4 - 4x^3$.

    Local max at $x = 0$; local min at $x = 3$. $f'(x) = 4x^3 - 12x^2$ gives critical points at $x = 0, 3$.

34. Identify the critical points of $f(x) = \frac{1}{x}$.

    No critical points; $f'(x)$ never zero. $f'(x) = -\frac{1}{x^2}$ is never zero, always negative.

35. Identify the local extrema of $f(x) = \frac{x^2}{2} - 2x$.

    Local min at $x = 2$. $f'(x) = x - 2 = 0$ when $x = 2$.

36. What is a necessary condition for local extrema?

    $f'(x) = 0$ or $f'(x)$ is undefined. Critical points are necessary but not sufficient conditions.

37. Determine local extrema for $f(x) = x^3 - 3x^2 + 4$.

    Local max at $x = 0$; local min at $x = 2$. $f'(x) = 3x^2 - 6x$ gives critical points at $x = 0, 2$.

38. What is a sufficient condition for a local maximum?

    $f'(x)$ changes from positive to negative. Derivative changes from positive to negative.

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