Concavity of Functions Over Their Domains - AP Calculus BC

1. Find $f''(x)$ for $f(x) = x^5$. What is the concavity at $x=1$?

   Concave up. $f''(x) = 20x^3$ and $f''(1) = 20 > 0$.

2. Identify the inflection point condition.

   $f''(x) = 0$ or $f''(x)$ changes sign. Where the second derivative is zero and changes sign.

3. What does $f''(x) < 0$ indicate about a function's graph?

   The graph is concave down. The curve bends downward like an upside-down bowl.

4. Evaluate $f''(x)$ for $f(x) = \text{ln}(x)$. What does it imply for $x > 0$?

   Concave down. $f''(x) = -\frac{1}{x^2} < 0$ for all $x > 0$.

5. Find $f''(x)$ for $f(x) = x^3 - 3x^2$ and determine concavity at $x=2$.

   Concave up. $f''(x) = 6x - 6$ and $f''(2) = 6 > 0$.

6. Analyze concavity of $f(x) = e^{-x}$ at $x = 0$.

   Concave up. $f''(x) = e^{-x} > 0$ for all $x$.

7. Determine the concavity for $f(x) = \frac{1}{x^2}$ for $x > 0$.

   Concave up. $f''(x) = \frac{6}{x^4} > 0$ for all $x ≠ 0$.

8. Is the function $f(x) = \text{ln}(x)$ concave up or down for $x > 0$?

   Concave down. $f''(x) = -\frac{1}{x^2} < 0$ for all $x > 0$.

9. What indicates a point of inflection in terms of $f''(x)$?

   $f''(x)$ changes sign at that point. The concavity switches direction at inflection points.

10. How do you determine concavity using the second derivative?

    Concave up if $f''(x) > 0$; concave down if $f''(x) < 0$. Positive second derivative means upward curve, negative means downward.

11. What does $f''(x) < 0$ throughout an interval imply?

    Function is concave down on that interval. The graph curves downward throughout the interval.

12. What is the second derivative test used for?

    To determine concavity and points of inflection. The second derivative reveals how the slope is changing.

13. State the general condition for a function to be concave up.

    $f''(x) > 0$ over the interval. When the second derivative is positive everywhere.

14. What does $f''(x) > 0$ throughout an interval imply?

    Function is concave up on that interval. The graph curves upward throughout the interval.

15. Determine the concavity of $f(x) = \frac{1}{3}x^3 + x$ at $x = 1$.

    Concave up. $f''(x) = 2x$ and $f''(1) = 2 > 0$.

16. Determine concavity for $f(x) = e^x$ for all $x$.

    Concave up. $f''(x) = e^x > 0$ for all real $x$.

17. Find $f''(x)$ for $f(x) = x^2 + 4x$. What does it imply for concavity?

    Concave up. $f''(x) = 2 > 0$, so always concave up.

18. What is the sign of $f''(x)$ when the function is concave down?

    $f''(x) < 0$. Negative second derivative indicates the graph curves downward.

19. Determine concavity for $f(x) = 3x^5 - 10x^3$ at $x = 2$.

    Concave up. $f''(x) = 60x^3 - 60x$ and $f''(2) = 360 > 0$.

20. What is the sign of $f''(x)$ when the function is concave up?

    $f''(x) > 0$. Positive second derivative indicates the graph curves upward.

21. Determine if $f(x) = 7x^2 - 3x$ is concave up or down at $x=0$.

    Concave up. $f''(x) = 14 > 0$, so always concave up.

22. Determine concavity for $f(x) = x^{2/3}$ at $x = 0$.

    Neither concave up nor down. $f''(0)$ is undefined; the function has a cusp.

23. Find the concavity for $f(x) = -x^2$ at $x = 0$.

    Concave down. $f''(x) = -2 < 0$, so the parabola opens downward.

24. Identify the concavity of $f(x) = x^7$ for $x < 0$.

    Concave down. $f''(x) = 42x^5 < 0$ when $x < 0$.

25. Determine the concavity of $f(x) = \text{cos}(x)$ at $x = 0$.

    Concave down. $f''(x) = -\cos(x)$ and $f''(0) = -1 < 0$.

26. For $f(x) = \text{sin}(x)$, identify concavity at $x = \frac{\text{pi}}{2}$.

    Concave down. $f''(x) = -\sin(x)$ and $f''(\frac{\pi}{2}) = -1 < 0$.

27. For $f(x) = x^3$, what does $f''(0)$ tell us?

    Possible inflection point. $f''(0) = 0$ and sign changes from negative to positive.

28. Find the concavity for $f(x) = x^2$ at $x = 0$.

    Concave up. $f''(x) = 2 > 0$, so the parabola opens upward.

29. Evaluate concavity for $f(x) = x^4 - 4x^3$ at $x = 3$.

    Concave up. $f''(x) = 12x^2 - 24x$ and $f''(3) = 36 > 0$.

30. Identify the concavity of $f(x) = x^4$ for $x > 0$.

    Concave up. $f''(x) = 12x^2 > 0$ for all $x ≠ 0$.

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