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

1. What is the definition of a continuous function?

   No gaps, jumps, or holes in domain. Function exists at every point with no discontinuities or breaks.

2. Is $f(x) = \frac{1}{x}$ on $[1, 3]$ continuous?

   Yes, $f(x)$ is continuous on $[1, 3]$. Rational functions are continuous where denominator is nonzero.

3. Identify the critical point for $f(x) = x^2$.

   Critical point: $x = 0$. $f'(x) = 2x = 0$ only when $x = 0$.

4. Find $f'(x)$ for $f(x) = \frac{x^2}{x+2}$.

   $f'(x) = \frac{2x(x+2)-x^2}{(x+2)^2}$. Quotient rule: $\frac{d}{dx}[\frac{x^2}{x+2}] = \frac{2x(x+2)-x^2}{(x+2)^2}$.

5. Describe how to find global extrema on $[a, b]$.

   Evaluate $f$ at critical points and endpoints. Compare function values at all candidate points to find extrema.

6. Is $f(x) = x^3$ on $[-1, 1]$ continuous?

   Yes, $f(x)$ is continuous on $[-1, 1]$. Polynomial functions are continuous everywhere on their domain.

7. Define a global maximum.

   $f(c) \geq f(x)$ for all $x$ in domain of $f$. Function value at $c$ exceeds all others in the entire domain.

8. What is the Second Derivative Test?

   Uses $f''(x)$ to determine concavity and local extrema. If $f''(c) > 0$, local min; if $f''(c) < 0$, local max.

9. Find $f'(x)$ for $f(x) = \frac{x}{x+1}$.

   $f'(x) = \frac{1}{(x+1)^2}$. Quotient rule applied: $\frac{d}{dx}[\frac{x}{x+1}] = \frac{1}{(x+1)^2}$.

10. What conditions are necessary for EVT to apply?

    Function must be continuous on closed interval. Closed interval ensures compactness for guaranteed extrema existence.

11. Identify the extrema of $f(x) = -x^2 + 4x - 3$ on $[0, 3]$.

    Global max at $x=2$, Global min at $x=3$. Vertex at $x=2$ gives max; endpoint at $x=3$ gives min.

12. What is a local maximum?

    $f(c) \geq f(x)$ for all $x$ near $c$. Function value at $c$ exceeds nearby values within some interval.

13. State Fermat's Theorem.

    If $f$ has local extremum at $c$, $f'(c)=0$ or undefined. Local extrema only occur where the derivative is zero or undefined.

14. What is the First Derivative Test?

    Determines local extrema using $f'(x)$ sign change. Analyzes where $f'$ changes from positive to negative or vice versa.

15. Identify global extrema of $f(x) = x^3 - 6x^2 + 9x + 1$ on $[0, 3]$.

    Global min at $x=0$, Global max at $x=3$. Evaluate at critical points and endpoints to compare values.

16. Find critical points of $f(x) = \frac{1}{3}x^3 - x^2$.

    Critical points: $x = 0, x = 2$. Set $f'(x) = x^2 - 2x = x(x-2) = 0$ to find solutions.

17. Find critical points of $f(x)=x^4-4x^2$.

    Critical points: $x = 0, x = 2, x = -2$. Set $f'(x) = 4x^3 - 8x = 4x(x^2-2) = 0$ to solve.

18. Find global extrema of $f(x) = 2x^2$ on $[-1, 2]$.

    Global min at $x=0$, Global max at $x=2$. Evaluate $f$ at critical point $x=0$ and endpoints $x=-1,2$.

19. Determine if $x=1$ is a critical point for $f(x) = x^3 - 3x$.

    Yes, $x=1$ is a critical point. $f'(1) = 3(1)^2 - 3 = 0$, so derivative equals zero.

20. Find global extrema for $f(x) = -x^2$ on $[-1, 1]$.

    Global max at $x=0$, Global min at $x=-1$ and $x=1$. Parabola opens downward with vertex at origin and symmetric endpoints.

21. Calculate $f'(x)$ for $f(x) = x^3 - 6x^2 + 9x$.

    $f'(x) = 3x^2 - 12x + 9$. Power rule applied: $\frac{d}{dx}[x^3 - 6x^2 + 9x] = 3x^2 - 12x + 9$.

22. Find the derivative of $f(x) = 3x^2 - 6x + 1$.

    $f'(x) = 6x - 6$. Power rule applied to each term of the polynomial.

23. What is the Extreme Value Theorem?

    If $f$ is continuous on $[a, b]$, it has a max and min. Guarantees absolute extrema exist on closed, bounded intervals.

24. What is a local minimum?

    $f(c) \leq f(x)$ for all $x$ near $c$. Function value at $c$ is below nearby values within some interval.

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

    Critical points: $x = 0, x = 2$. Set $f'(x) = 3x^2 - 6x = 3x(x-2) = 0$ to find critical points.

26. What is the derivative test for local extrema?

    Use $f'(x)$ sign changes to identify local extrema. Sign changes in $f'(x)$ indicate transitions between increasing/decreasing.

27. Find critical points of $f(x) = x^2 - 1$.

    Critical point: $x = 0$. $f'(x) = 2x = 0$ only at $x = 0$.

28. Determine if $x=0$ is a critical point for $f(x)=x^4$.

    Yes, $x=0$ is a critical point. $f'(0) = 4(0)^3 = 0$, so derivative equals zero at origin.

29. Define a global minimum.

    $f(c) \leq f(x)$ for all $x$ in domain of $f$. Function value at $c$ is below all others in the entire domain.

30. Find critical points of $f(x) = x^2 - 4x$.

    Critical points: $x = 2$. Set $f'(x) = 2x - 4 = 0$ to solve for critical points.

31. Identify extrema of $f(x) = x^2 - 2x + 1$ on $[0, 2]$.

    Global min at $x=1$, Global max at $x=2$. Vertex of parabola at $x=1$ gives min; endpoint gives max.

32. What is a critical point?

    Where $f'(x) = 0$ or $f'(x)$ is undefined. Potential locations for local extrema based on derivative behavior.

33. Identify the global extrema of $f(x) = x^2 - 4x$ on $[0, 3]$.

    Global min at $x=2$, Global max at $x=3$. Vertex at $x=2$ gives min; endpoint at $x=3$ gives max.

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