Candidates Test - AP Calculus AB

  1. Find f(x)f(x) at x=5x = 5 for f(x)=x2−5xf(x) = x^2 - 5x.
    f(5)=0f(5) = 0. Substitution shows x=5x=5 is a root of the function.
  2. What is the Candidates Test used for in calculus?
    Determining the absolute extrema on a closed interval. Finds global max/min on closed intervals by testing all candidate points.
  3. What are the critical points of f(x)=x2−9x+20f(x) = x^2 - 9x + 20?
    Critical points: x=92x = \frac{9}{2}. Set f′(x)=2x−9=0f'(x)=2x-9=0 and solve for xx.
  4. Identify the type of interval required for the Candidates Test.
    A closed interval [a,b][a, b]. Ensures function is continuous and extrema exist by EVT.
  5. Find f(x)f(x) at x=0x = 0 for f(x)=x4−2x2f(x) = x^4 - 2x^2.
    f(0)=0f(0) = 0. Direct substitution of x=0x=0 into the function.
  6. If f′(x)f'(x) changes sign at x=cx = c, what is x=cx = c?
    A critical point. Sign changes in derivative indicate critical points.
  7. State the first step in the Candidates Test.
    Find the derivative of the function. Need derivative to locate critical points where extrema may occur.
  8. Which function value determines the absolute maximum?
    The largest value among evaluated points. Highest function output among all tested candidates.
  9. Find f(1)f(1) for f(x)=x2−2x+1f(x) = x^2 - 2x + 1.
    f(1)=0f(1) = 0. This is a perfect square: (x−1)2=0(x-1)^2=0 at x=1x=1.
  10. Find the derivative of f(x)=x3−3x2+4f(x) = x^3 - 3x^2 + 4.
    f′(x)=3x2−6xf'(x) = 3x^2 - 6x. Apply power rule: ddx[x3]=3x2\frac{d}{dx}[x^3]=3x^2, etc.
  11. What is a critical point?
    Where f′(x)=0f'(x) = 0 or f′(x)f'(x) is undefined. Points where derivative equals zero or doesn't exist.
  12. Evaluate f′(x)=4x−4f'(x) = 4x - 4 for critical points.
    Critical point: x=1x = 1. Set 4x−4=04x-4=0 and solve for xx.
  13. Determine the endpoints for the interval [a,b]=[−2,5][a, b] = [-2, 5].
    Endpoints are x=−2x = -2 and x=5x = 5. These are the boundary values of the given interval.
  14. Evaluate f(x)f(x) at endpoints for f(x)=x2−2xf(x) = x^2 - 2x on [0,3][0, 3].
    f(0)=0f(0) = 0, f(3)=3f(3) = 3. Substitute boundary values into the function.
  15. What is the final step in the Candidates Test?
    Compare function values at critical points and endpoints. Determines which candidate gives the absolute maximum/minimum.
  16. State the condition for a point to be a candidate for extrema.
    Point must be a critical point or endpoint. Only these points can produce absolute extrema on closed intervals.
  17. Find f(x)f(x) at x=3x = 3 for f(x)=x3−6x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1.
    f(3)=1f(3) = 1. Substitute x=3x=3 into the function.
  18. Evaluate f′(x)=3x2−12x+9f'(x) = 3x^2 - 12x + 9 for critical points.
    Critical points: x=1,x=3x = 1, x = 3. Solve 3x2−12x+9=03x^2-12x+9=0 to get x=1,3x=1,3.
  19. Identify the formula for finding critical points.
    Set f′(x)=0f'(x) = 0 and solve for xx. Standard method to find where slope equals zero.
  20. Evaluate f′(x)=6x−6f'(x) = 6x - 6 for critical points.
    Critical point: x=1x = 1. Set 6x−6=06x-6=0 and solve for xx.
  21. Which values are evaluated in the Candidates Test?
    Endpoints and critical points of the interval. All possible locations where absolute extrema can occur.
  22. Find f(x)f(x) at x=2x = 2 for f(x)=x2−4x+4f(x) = x^2 - 4x + 4.
    f(2)=0f(2) = 0. Direct substitution shows this is a perfect square.
  23. Determine f(x)f(x) at x=2x=2 for f(x)=x3−3x2+4xf(x) = x^3 - 3x^2 + 4x.
    f(2)=4f(2) = 4. Direct substitution of x=2x=2 into the cubic function.
  24. Evaluate f(x)f(x) at endpoints for f(x)=1xf(x) = \frac{1}{x} on [1,3][1, 3].
    f(1)=1f(1) = 1, f(3)=13f(3) = \frac{1}{3}. Substitute endpoint values into f(x)=1xf(x)=\frac{1}{x}.
  25. What is the derivative of f(x)=x3−6x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1?
    f′(x)=3x2−12x+9f'(x) = 3x^2 - 12x + 9. Apply power rule to each term.
  26. What is f(x)f(x) at x=0x = 0 for f(x)=1x+xf(x) = \frac{1}{x} + x on [1,2][1, 2]?
    Not applicable; x=0x = 0 is not in [1,2][1, 2]. Point outside domain cannot be evaluated.
  27. What must be determined after finding the derivative in the Candidates Test?
    Critical points where the derivative is zero or undefined. These are locations where extrema can occur on the interval.
  28. Which function value determines the absolute minimum?
    The smallest value among evaluated points. Lowest function output among all tested candidates.
  29. What are endpoints in the context of the Candidates Test?
    The boundary values of the closed interval. The left and right limits of the domain interval.

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