Candidates Test - AP Calculus BC

1. Which theorem supports the use of the Candidates Test?

   Extreme Value Theorem. Guarantees continuous functions have absolute extrema on closed intervals.

2. Evaluate $f(x) = -x^3 + 3x^2$ at endpoints $x = 0$ and $x = 2$.

   $f(0) = 0$, $f(2) = 0$. Shows function values at the boundary points of the interval.

3. Evaluate $f(x) = x^2$ at $x = -1, 0, 1$ to find extrema.

   Minimum at $x = 0$, Maximum at $x = -1$ or $x = 1$. $f(-1) = 1$, $f(0) = 0$, $f(1) = 1$, so min at $x = 0$, max at endpoints.

4. What is the final step in the Candidates Test?

   Compare function values to identify absolute extrema. The highest and lowest function values give the absolute max and min.

5. What is the absolute maximum of $f(x) = x^4 - 4x^2 + 4$ on $[-2, 2]$?

   Maximum at $x = 0$. Critical points at $x = \pm\sqrt{2}$ give local minima, max at $x = 0$.

6. Evaluate $f(x) = x^2 - 4x + 3$ at endpoints $x = 0$ and $x = 3$.

   $f(0) = 3$, $f(3) = -3$. Direct substitution into the function at the boundary points.

7. Evaluate $f(x) = x^2 + x - 6$ at $x = -3, 0, 2$.

   $f(-3) = 0$, $f(0) = -6$, $f(2) = 0$. Direct evaluation by substituting each x-value into the function.

8. Determine the absolute minimum of $f(x) = x^2$ on $[-1, 1]$.

   Minimum at $x = 0$. Critical point $x = 0$ gives $f(0) = 0$, the minimum value.

9. What is a critical point?

   Point where $f'(x) = 0$ or $f'(x)$ is undefined. These are potential locations for local extrema of the function.

10. State the condition for using limits in the Candidates Test.

    Limits are used if endpoints are undefined. When function is undefined at endpoints, use limit values instead.

11. How do you find critical points?

    Solve $f'(x) = 0$ and check where $f'(x)$ is undefined. Both conditions identify all points where extrema can occur.

12. What is the Extreme Value Theorem?

    A continuous function on a closed interval has absolute extrema. The mathematical foundation that justifies the Candidates Test method.

13. Identify the second step in the Candidates Test.

    Evaluate the function at critical points and endpoints. All candidate points must be tested to find absolute extrema.

14. What is the absolute minimum of $f(x) = 1 - x^2$ on $[-1, 1]$?

    -1 at $x = -1$ or $x = 1$. Parabola opens downward with maximum at $x = 0$, minimum at endpoints.

15. Identify the absolute extrema for $f(x) = \frac{1}{x^2}$ on $[1, 3]$.

    Minimum at $x = 3$, Maximum at $x = 1$. $f(x) = \frac{1}{x^2}$ decreases on $[1,3]$, so max at $x=1$, min at $x=3$.

16. Why is continuity important for the Candidates Test?

    Ensures extrema exist on closed intervals. Without continuity, absolute extrema might not exist.

17. State the first step in the Candidates Test.

    Find the derivative and solve for critical points. Critical points occur where the derivative equals zero or is undefined.

18. What is the absolute maximum of $f(x) = x^3$ on $[-1, 2]$?

    $f(2) = 8$. Evaluating at endpoints and critical point shows maximum at $x = 2$.

19. What is the Candidates Test used for?

    Determining absolute extrema on a closed interval. Finds global max/min on closed intervals using critical points and endpoints.

20. Find the critical points for $f(x) = x^3 - 3x^2 + 4$.

    $x = 0, x = 2$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, so $x = 0$ and $x = 2$.

21. Evaluate $f(x) = 3x^2 - 12x$ at $x = 0, 2, 4$.

    $f(0) = 0$, $f(2) = -12$, $f(4) = 0$. Direct substitution shows function values at these candidate points.

22. How is the derivative used in the Candidates Test?

    To find critical points. Sets $f'(x) = 0$ to locate potential extrema within the interval.

23. Identify the absolute extrema for $f(x) = x^2 - 4$ on $[-3, 3]$.

    Maximum at $x = 3$, Minimum at $x = 0$. Compare $f(-3) = 5$, $f(0) = -4$, $f(3) = 5$ for extrema locations.

24. What is the absolute maximum of $f(x) = -x^2 + 4$ on $[-2, 2]$?

    4 at $x = 0$. Critical point at $x = 0$ gives maximum value since $f'(x) = -2x$.

25. What is the purpose of evaluating endpoints in the Candidates Test?

    To ensure all potential extrema are considered. Extrema can occur at boundary points of the closed interval.

26. Find the absolute minimum of $f(x) = x^3 - 3x$ on $[-2, 2]$.

    Minimum at $x = -2$. Critical point $x = 1$ gives local max, compare with endpoint values.

27. Determine the absolute minimum of $f(x) = x^3 - 3x^2 + 2$ on $[0, 3]$.

    Minimum at $x = 2$. Critical point $x = 2$ gives $f(2) = -2$, the lowest value on the interval.

28. What role do endpoints play in the Candidates Test?

    They are evaluated for potential extrema. Include boundary points as candidates since extrema can occur there.

29. Find the absolute extrema for $f(x) = \frac{1}{x}$ on $[1, 4]$.

    Minimum at $x = 4$, Maximum at $x = 1$. $f(x) = \frac{1}{x}$ decreases on $[1,4]$, so max at $x=1$, min at $x=4$.

30. Evaluate $f(x) = x^2 + 1$ at $x = -2, 0, 2$.

    $f(-2) = 5$, $f(0) = 1$, $f(2) = 5$. Shows function values at critical point and endpoints for comparison.

31. What function values are compared in the Candidates Test?

    Values at critical points and endpoints. These are all possible locations where absolute extrema can occur.

32. Find the absolute extrema of $f(x) = |x|$ on $[-2, 2]$.

    Minimum at $x = 0$, Maximum at $x = -2$ or $x = 2$. Critical point at $x = 0$ gives minimum, endpoints give maximum values.

33. What condition must a function meet to apply the Candidates Test?

    Function must be continuous on a closed interval. Ensures the function has guaranteed absolute maximum and minimum values.

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