Second Derivative Test - AP Calculus BC

1. What is the second derivative of $f(x) = -x^4 + 2x^2 - x$?

   $f''(x) = -12x^2 + 4$. Differentiate $f'(x) = -4x^3 + 4x - 1$ to get the second derivative.

2. What does $f''(c) < 0$ imply about $f(x)$ at $x=c$?

   Local maximum. Negative second derivative confirms the critical point is a maximum.

3. What is the sign of $f''(x)$ if $f(x)$ is concave up?

   Positive. Concave up functions have positive curvature everywhere.

4. What is the second derivative of $f(x) = 3x^4 - 8x^2 + 5$?

   $f''(x) = 36x^2 - 16$. Differentiate $f'(x) = 12x^3 - 16x$ to get the second derivative.

5. What does $f''(c) = 0$ indicate about the critical point $c$?

   Test is inconclusive at $c$. Zero second derivative provides no information about the extremum type.

6. Determine the nature of the extremum at $x = \frac{1}{2}$ for $f(x) = -x^2 + x$.

   Local maximum. $f''(1/2) = -2 < 0$ at the critical point, indicating a maximum.

7. Determine if $f(x) = x^3 - 3x^2 + 4$ has a local extremum at $x = 2$.

   Local minimum at $x = 2$. $f'(2) = 0$ and $f''(2) = 6 > 0$, confirming a local minimum.

8. What does $f''(c) > 0$ imply about $f(x)$ at $x=c$?

   Local minimum. Positive second derivative confirms the critical point is a minimum.

9. Evaluate $f''(1)$ for $f(x) = x^4 - x^2 + x$.

   $f''(1) = 10$. $f'(x) = 4x^3 - 2x + 1$ and $f''(x) = 12x^2 - 2$, so $f''(1) = 10$.

10. What is the sign of $f''(x)$ if $f(x)$ is concave down?

    Negative. Concave down functions have negative curvature everywhere.

11. Find $f''(x)$ for $f(x) = x^4 - 4x^2 + 2$.

    $f''(x) = 12x^2 - 8$. Take the derivative of $f'(x) = 4x^3 - 8x$ to get the second derivative.

12. What is the second derivative test used for?

    Determining local extrema of a function. Uses concavity at critical points to classify local maxima and minima.

13. Determine the nature of the extremum for $f(x) = x^2 - 4x + 4$ at $x = 2$.

    Local minimum. $f'(2) = 0$ and $f''(2) = 2 > 0$, confirming a local minimum.

14. What conclusion is drawn if $f''(c) = 0$ at a critical point $c$?

    Test inconclusive. Zero second derivative means the test cannot determine extremum type.

15. Calculate $f''(3)$ for $f(x) = x^3 - 9x^2 + 27x - 3$.

    $f''(3) = 0$. $f'(x) = 3x^2 - 18x + 27$ and $f''(x) = 6x - 18$, so $f''(3) = 0$.

16. Identify the sign of $f''(x)$ for a concave down interval.

    $f''(x) < 0$. Negative second derivative always indicates downward concavity.

17. Identify the sign of $f''(x)$ for a concave up interval.

    $f''(x) > 0$. Positive second derivative always indicates upward concavity.

18. Find the second derivative of $f(x) = \frac{1}{3}x^3 - 2x^2 + 3x$.

    $f''(x) = 2x - 4$. $f'(x) = x^2 - 4x + 3$, so $f''(x) = 2x - 4$.

19. Determine the nature of the extremum for $f(x) = x^2 - 4$ at $x = 0$.

    Local minimum. $f'(0) = 0$ and $f''(0) = 2 > 0$, confirming a local minimum.

20. What is the second derivative of $f(x) = 2x^3 - 3x^2 + x$?

    $f''(x) = 12x - 6$. Differentiate $f'(x) = 6x^2 - 6x + 1$ to get the second derivative.

21. What is the second derivative of $f(x) = -x^3 + 3x - 5$?

    $f''(x) = -6x$. Differentiate $f'(x) = -3x^2 + 3$ to get the second derivative.

22. Find the second derivative of $f(x) = x^5 - 5x^3 + 7$.

    $f''(x) = 20x^3 - 30x$. Differentiate $f'(x) = 5x^4 - 15x^2$ to get the second derivative.

23. State the second derivative test condition for a local minimum.

    $f''(c) > 0$ at critical point $c$. Positive second derivative indicates concave up, confirming a minimum.

24. Determine concavity for $f(x) = 2x^2 - 3x + 1$ at $x = 0$.

    Concave up. $f''(x) = 4 > 0$ everywhere, so the function is concave up.

25. Evaluate $f''(0)$ for $f(x) = x^4 - 4x^2 + 1$.

    $f''(0) = -8$. $f'(x) = 4x^3 - 8x$, so $f''(x) = 12x^2 - 8$ and $f''(0) = -8$.

26. At which type of point is the second derivative test applied?

    Critical points. Only points where $f'(x) = 0$ can be tested for extrema.

27. What is the condition for test inconclusiveness in the second derivative test?

    $f''(c) = 0$ at $c$. This condition makes the second derivative test unable to determine extremum type.

28. Evaluate $f''(2)$ for $f(x) = x^4 - 4x^3 + 6x^2$.

    $f''(2) = 12$. $f'(x) = 4x^3 - 12x^2 + 12x$, so $f''(x) = 12x^2 - 24x + 12$.

29. State the second derivative test condition for a local maximum.

    $f''(c) < 0$ at critical point $c$. Negative second derivative indicates concave down, confirming a maximum.

30. Find the second derivative of $f(x) = x^3 + 3x^2 + x + 1$.

    $f''(x) = 6x + 6$. Differentiate $f'(x) = 3x^2 + 6x + 1$ to get the second derivative.

31. Identify the nature of the critical point for $f(x) = x^3 - 3x$ at $x = 0$.

    Inconclusive. $f''(0) = 0$ at this critical point, so the test fails.

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