Working With the Intermediate Value Theorem - AP Calculus BC

1. Identify the key feature of a function for IVT applicability.

   Continuity on the interval $[a, b]$. No gaps or jumps ensure all intermediate values between endpoints exist.

2. Is $N = 0$ reachable for $f(x) = x^2 - 1$ on $[0, 2]$?

   Yes, because $f(0) < 0 < f(2)$. $f(0) = -1 < 0 < 3 = f(2)$ confirms IVT applies for reaching zero.

3. Does $f(x) = \text{cos}(x)$ satisfy IVT on $[0, \frac{3\text{π}}{2}]$, $N = 0$?

   Yes, because $f(0) > 0 > f(\frac{3\text{π}}{2})$. $f(0) = 1 > 0 > -1 = f(\frac{3\pi}{2})$ confirms IVT applies for zero.

4. Find if IVT applies: $f(x) = \text{ln}(x)$ on $(0, 1]$, $N = -1$.

   No, IVT requires closed interval $[0, 1]$. Open interval $(0,1]$ violates IVT requirement for closed intervals.

5. Apply IVT: $f(x) = \frac{1}{x}$ on $[1, 3]$, $N = 0.5$.

   Yes, because $f$ is continuous and $f(1) > 0.5 > f(3)$. $f(1) = 1$ and $f(3) = \frac{1}{3}$, so $\frac{1}{3} < 0.5 < 1$.

6. Check IVT for $f(x) = x^3 - x$ on $[0, 2]$, $N = 0.5$.

   Yes, $f(0) < 0.5 < f(2)$. $f(0) = 0$ and $f(2) = 6$, so $0 < 0.5 < 6$ with continuity.

7. Does $f(x) = x^2 + 1$ satisfy IVT on $[-1, 1]$, $N = 0$?

   No, because $f(x) \text{ is never } 0$. Function has minimum value 1, so zero is not in the range.

8. Check if $f(x) = x^4 - 2x^2$ satisfies IVT on $[0, 2]$, $N = 1$.

   Yes, because $f(0) < 1 < f(2)$. $f(0) = 0 < 1 < 8 = f(2)$ confirms IVT guarantees the value exists.

9. Check IVT for $f(x) = x^2 - 5$ on $[1, 3]$, $N = 3$.

   Yes, because $f(1) < 3 < f(3)$. $f(1) = -4 < 3 < 4 = f(3)$ confirms IVT applies for $N = 3$.

10. What does the IVT not guarantee?

    The specific location of $c$ where $f(c) = N$. IVT only guarantees existence, not the exact location of the solution.

11. Identify if IVT can be applied: $f(x) = 2x + 1$ on $[-1, 1]$, $N = 0$.

    Yes, because $f(-1) < 0 < f(1)$. $f(-1) = -1 < 0 < 3 = f(1)$ confirms IVT applies for zero.

12. Does the IVT apply to $f(x) = 1/x$ on $[-1, 1]$?

    No, because $f(x)$ is not continuous on $[-1, 1]$. Division by zero at $x = 0$ creates discontinuity within the interval.

13. Evaluate IVT applicability: $f(x) = \text{sin}(x)$ on $[0, \frac{\text{π}}{6}]$, $N = 0.3$.

    Yes, $f(0) < 0.3 < f(\frac{\text{π}}{6})$. $f(0) = 0 < 0.3 < 0.5 = f(\frac{\pi}{6})$ confirms IVT applies.

14. Find if IVT applies: $f(x) = e^x$ on $[0, 1]$, $N = 2$.

    Yes, $f(0) < 2 < f(1)$, so IVT applies. $f(0) = 1$ and $f(1) = e \approx 2.718$, so $1 < 2 < e$.

15. Does $f(x) = x^2 - 4$ have a root in $[1, 3]$ by IVT?

    Yes, because $f(1) < 0 < f(3)$. $f(1) = -3 < 0 < 5 = f(3)$ confirms a root exists by IVT.

16. Verify IVT for $f(x) = x^3 + x$ on $[0, 1]$, $N = 0.5$.

    Yes, $f(0) < 0.5 < f(1)$. $f(0) = 0 < 0.5 < 2 = f(1)$ confirms IVT applies with continuity.

17. State the conclusion of the IVT for $f(x) = x^3$ on $[1, 9]$, $N = 5$.

    There exists $c \text{ in } (1, 9)$ such that $f(c) = 5$. $f(1) = 1 < 5 < 729 = f(9)$, so IVT guarantees a solution exists.

18. What is required for a value $N$ in IVT?

    $N$ must be between $f(a)$ and $f(b)$. $N$ must lie between the function values at the interval's endpoints.

19. Evaluate if IVT holds for $f(x) = x^2 - 2$ on $[1, 2]$, $N = 0.5$.

    Yes, $f(1) < 0.5 < f(2)$. $f(1) = -1 < 0.5 < 2 = f(2)$ confirms IVT applies for the value.

20. Identify if the IVT can be applied: $f(x) = x^2$, interval $[-1, 2]$, $N = 3$.

    Yes, because $f$ is continuous and $f(-1) < 3 < f(2)$. $f(-1) = 1$ and $f(2) = 4$, so $1 < 3 < 4$ with continuity satisfied.

21. What condition must a function meet to apply the IVT?

    The function must be continuous on the closed interval $[a, b]$. Continuity ensures no gaps or jumps that could skip intermediate values.

22. What does the IVT guarantee about $f(c)$ if $f(a) < N < f(b)$?

    There exists $c \text{ in } (a, b)$ such that $f(c) = N$. IVT guarantees existence of $c$ where the function equals the intermediate value.

23. Determine if IVT holds for $f(x) = \text{sin}(x)$ on $[0, \frac{\text{π}}{2}]$, $N = 0.5$.

    Yes, because $f(0) < 0.5 < f\big(\frac{\text{π}}{2}\big)$. $f(0) = 0$ and $f(\frac{\pi}{2}) = 1$, so $0 < 0.5 < 1$ with continuity.

24. Can IVT be applied to $f(x) = \frac{1}{x-1}$ on $[0, 2]$?

    No, $f(x)$ is not continuous on $[0, 2]$. Vertical asymptote at $x = 1$ creates discontinuity within the interval.

25. Which type of intervals does the IVT require?

    Closed intervals $[a, b]$. Closed intervals include endpoints where continuity must be established.

26. Determine IVT applicability: $f(x) = x^2 - 4$ on $[2, 4]$, $N = 0$.

    No, $f(x)$ is never $0$ on $[2, 4]$. Function equals zero at $x = \pm 2$, but interval $[2,4]$ has all positive values.

27. Determine if IVT applies: $f(x) = x(x-1)$ on $[0, 2]$, $N = 0.5$.

    Yes, $f(0) < 0.5 < f(2)$. $f(0) = 0 < 0.5 < 2 = f(2)$ confirms IVT guarantees the value exists.

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