Working With the Intermediate Value Theorem - AP Calculus BC
Card 1 of 30
Is $N = 0$ reachable for $f(x) = x^2 - 1$ on $[0, 2]$?
Is $N = 0$ reachable for $f(x) = x^2 - 1$ on $[0, 2]$?
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Yes, because $f(0) < 0 < f(2)$. $f(0) = -1 < 0 < 3 = f(2)$ confirms IVT applies for reaching zero.
Yes, because $f(0) < 0 < f(2)$. $f(0) = -1 < 0 < 3 = f(2)$ confirms IVT applies for reaching zero.
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Does $f(x) = x^2 - 4$ have a root in $[1, 3]$ by IVT?
Does $f(x) = x^2 - 4$ have a root in $[1, 3]$ by IVT?
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Yes, because $f(1) < 0 < f(3)$. $f(1) = -3 < 0 < 5 = f(3)$ confirms a root exists by IVT.
Yes, because $f(1) < 0 < f(3)$. $f(1) = -3 < 0 < 5 = f(3)$ confirms a root exists by IVT.
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Can IVT be applied to $f(x) = \frac{1}{x-1}$ on $[0, 2]$?
Can IVT be applied to $f(x) = \frac{1}{x-1}$ on $[0, 2]$?
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No, $f(x)$ is not continuous on $[0, 2]$. Vertical asymptote at $x = 1$ creates discontinuity within the interval.
No, $f(x)$ is not continuous on $[0, 2]$. Vertical asymptote at $x = 1$ creates discontinuity within the interval.
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Determine if IVT holds for $f(x) = \text{sin}(x)$ on $[0, \frac{\text{π}}{2}]$, $N = 0.5$.
Determine if IVT holds for $f(x) = \text{sin}(x)$ on $[0, \frac{\text{π}}{2}]$, $N = 0.5$.
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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.
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.
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Apply IVT: $f(x) = \frac{1}{x}$ on $[1, 3]$, $N = 0.5$.
Apply IVT: $f(x) = \frac{1}{x}$ on $[1, 3]$, $N = 0.5$.
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Yes, because $f$ is continuous and $f(1) > 0.5 > f(3)$. $f(1) = 1$ and $f(3) = \frac{1}{3}$, so $\frac{1}{3} < 0.5 < 1$.
Yes, because $f$ is continuous and $f(1) > 0.5 > f(3)$. $f(1) = 1$ and $f(3) = \frac{1}{3}$, so $\frac{1}{3} < 0.5 < 1$.
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Find if IVT applies: $f(x) = e^x$ on $[0, 1]$, $N = 2$.
Find if IVT applies: $f(x) = e^x$ on $[0, 1]$, $N = 2$.
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Yes, $f(0) < 2 < f(1)$, so IVT applies. $f(0) = 1$ and $f(1) = e \approx 2.718$, so $1 < 2 < e$.
Yes, $f(0) < 2 < f(1)$, so IVT applies. $f(0) = 1$ and $f(1) = e \approx 2.718$, so $1 < 2 < e$.
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Find if IVT applies: $f(x) = \text{ln}(x)$ on $(0, 1]$, $N = -1$.
Find if IVT applies: $f(x) = \text{ln}(x)$ on $(0, 1]$, $N = -1$.
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No, IVT requires closed interval $[0, 1]$. Open interval $(0,1]$ violates IVT requirement for closed intervals.
No, IVT requires closed interval $[0, 1]$. Open interval $(0,1]$ violates IVT requirement for closed intervals.
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What does the IVT guarantee about $f(c)$ if $f(a) < N < f(b)$?
What does the IVT guarantee about $f(c)$ if $f(a) < N < f(b)$?
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There exists $c \text{ in } (a, b)$ such that $f(c) = N$. IVT guarantees existence of $c$ where the function equals the intermediate value.
There exists $c \text{ in } (a, b)$ such that $f(c) = N$. IVT guarantees existence of $c$ where the function equals the intermediate value.
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Does the IVT apply to $f(x) = 1/x$ on $[-1, 1]$?
Does the IVT apply to $f(x) = 1/x$ on $[-1, 1]$?
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No, because $f(x)$ is not continuous on $[-1, 1]$. Division by zero at $x = 0$ creates discontinuity within the interval.
No, because $f(x)$ is not continuous on $[-1, 1]$. Division by zero at $x = 0$ creates discontinuity within the interval.
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What condition must a function meet to apply the IVT?
What condition must a function meet to apply the IVT?
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The function must be continuous on the closed interval $[a, b]$. Continuity ensures no gaps or jumps that could skip intermediate values.
The function must be continuous on the closed interval $[a, b]$. Continuity ensures no gaps or jumps that could skip intermediate values.
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Evaluate IVT applicability: $f(x) = \text{sin}(x)$ on $[0, \frac{\text{π}}{6}]$, $N = 0.3$.
Evaluate IVT applicability: $f(x) = \text{sin}(x)$ on $[0, \frac{\text{π}}{6}]$, $N = 0.3$.
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Yes, $f(0) < 0.3 < f(\frac{\text{π}}{6})$. $f(0) = 0 < 0.3 < 0.5 = f(\frac{\pi}{6})$ confirms IVT applies.
Yes, $f(0) < 0.3 < f(\frac{\text{π}}{6})$. $f(0) = 0 < 0.3 < 0.5 = f(\frac{\pi}{6})$ confirms IVT applies.
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What does the IVT not guarantee?
What does the IVT not guarantee?
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The specific location of $c$ where $f(c) = N$. IVT only guarantees existence, not the exact location of the solution.
The specific location of $c$ where $f(c) = N$. IVT only guarantees existence, not the exact location of the solution.
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What is required for a value $N$ in IVT?
What is required for a value $N$ in IVT?
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$N$ must be between $f(a)$ and $f(b)$. $N$ must lie between the function values at the interval's endpoints.
$N$ must be between $f(a)$ and $f(b)$. $N$ must lie between the function values at the interval's endpoints.
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Identify if the IVT can be applied: $f(x) = x^2$, interval $[-1, 2]$, $N = 3$.
Identify if the IVT can be applied: $f(x) = x^2$, interval $[-1, 2]$, $N = 3$.
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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.
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.
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State the conclusion of the IVT for $f(x) = x^3$ on $[1, 9]$, $N = 5$.
State the conclusion of the IVT for $f(x) = x^3$ on $[1, 9]$, $N = 5$.
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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.
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.
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Determine if IVT applies: $f(x) = x(x-1)$ on $[0, 2]$, $N = 0.5$.
Determine if IVT applies: $f(x) = x(x-1)$ on $[0, 2]$, $N = 0.5$.
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Yes, $f(0) < 0.5 < f(2)$. $f(0) = 0 < 0.5 < 2 = f(2)$ confirms IVT guarantees the value exists.
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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Which type of intervals does the IVT require?
Which type of intervals does the IVT require?
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Closed intervals $[a, b]$. Closed intervals include endpoints where continuity must be established.
Closed intervals $[a, b]$. Closed intervals include endpoints where continuity must be established.
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Check if $f(x) = x^4 - 2x^2$ satisfies IVT on $[0, 2]$, $N = 1$.
Check if $f(x) = x^4 - 2x^2$ satisfies IVT on $[0, 2]$, $N = 1$.
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Yes, because $f(0) < 1 < f(2)$. $f(0) = 0 < 1 < 8 = f(2)$ confirms IVT guarantees the value exists.
Yes, because $f(0) < 1 < f(2)$. $f(0) = 0 < 1 < 8 = f(2)$ confirms IVT guarantees the value exists.
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Identify if IVT can be applied: $f(x) = 2x + 1$ on $[-1, 1]$, $N = 0$.
Identify if IVT can be applied: $f(x) = 2x + 1$ on $[-1, 1]$, $N = 0$.
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Yes, because $f(-1) < 0 < f(1)$. $f(-1) = -1 < 0 < 3 = f(1)$ confirms IVT applies for zero.
Yes, because $f(-1) < 0 < f(1)$. $f(-1) = -1 < 0 < 3 = f(1)$ confirms IVT applies for zero.
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Check IVT for $f(x) = x^3 - x$ on $[0, 2]$, $N = 0.5$.
Check IVT for $f(x) = x^3 - x$ on $[0, 2]$, $N = 0.5$.
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Yes, $f(0) < 0.5 < f(2)$. $f(0) = 0$ and $f(2) = 6$, so $0 < 0.5 < 6$ with continuity.
Yes, $f(0) < 0.5 < f(2)$. $f(0) = 0$ and $f(2) = 6$, so $0 < 0.5 < 6$ with continuity.
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Evaluate if IVT holds for $f(x) = x^2 - 2$ on $[1, 2]$, $N = 0.5$.
Evaluate if IVT holds for $f(x) = x^2 - 2$ on $[1, 2]$, $N = 0.5$.
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Yes, $f(1) < 0.5 < f(2)$. $f(1) = -1 < 0.5 < 2 = f(2)$ confirms IVT applies for the value.
Yes, $f(1) < 0.5 < f(2)$. $f(1) = -1 < 0.5 < 2 = f(2)$ confirms IVT applies for the value.
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Determine IVT applicability: $f(x) = x^2 - 4$ on $[2, 4]$, $N = 0$.
Determine IVT applicability: $f(x) = x^2 - 4$ on $[2, 4]$, $N = 0$.
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No, $f(x)$ is never $0$ on $[2, 4]$. Function equals zero at $x = \pm 2$, but interval $[2,4]$ has all positive values.
No, $f(x)$ is never $0$ on $[2, 4]$. Function equals zero at $x = \pm 2$, but interval $[2,4]$ has all positive values.
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Verify IVT for $f(x) = x^3 + x$ on $[0, 1]$, $N = 0.5$.
Verify IVT for $f(x) = x^3 + x$ on $[0, 1]$, $N = 0.5$.
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Yes, $f(0) < 0.5 < f(1)$. $f(0) = 0 < 0.5 < 2 = f(1)$ confirms IVT applies with continuity.
Yes, $f(0) < 0.5 < f(1)$. $f(0) = 0 < 0.5 < 2 = f(1)$ confirms IVT applies with continuity.
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Does $f(x) = \text{cos}(x)$ satisfy IVT on $[0, \frac{3\text{π}}{2}]$, $N = 0$?
Does $f(x) = \text{cos}(x)$ satisfy IVT on $[0, \frac{3\text{π}}{2}]$, $N = 0$?
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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.
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.
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Check IVT for $f(x) = x^2 - 5$ on $[1, 3]$, $N = 3$.
Check IVT for $f(x) = x^2 - 5$ on $[1, 3]$, $N = 3$.
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Yes, because $f(1) < 3 < f(3)$. $f(1) = -4 < 3 < 4 = f(3)$ confirms IVT applies for $N = 3$.
Yes, because $f(1) < 3 < f(3)$. $f(1) = -4 < 3 < 4 = f(3)$ confirms IVT applies for $N = 3$.
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Identify the key feature of a function for IVT applicability.
Identify the key feature of a function for IVT applicability.
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Continuity on the interval $[a, b]$. No gaps or jumps ensure all intermediate values between endpoints exist.
Continuity on the interval $[a, b]$. No gaps or jumps ensure all intermediate values between endpoints exist.
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Does the IVT apply to $f(x) = 1/x$ on $[-1, 1]$?
Does the IVT apply to $f(x) = 1/x$ on $[-1, 1]$?
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No, because $f(x)$ is not continuous on $[-1, 1]$. Division by zero at $x = 0$ creates discontinuity within the interval.
No, because $f(x)$ is not continuous on $[-1, 1]$. Division by zero at $x = 0$ creates discontinuity within the interval.
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Determine IVT applicability: $f(x) = x^2 - 4$ on $[2, 4]$, $N = 0$.
Determine IVT applicability: $f(x) = x^2 - 4$ on $[2, 4]$, $N = 0$.
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No, $f(x)$ is never $0$ on $[2, 4]$. Function equals zero at $x = \pm 2$, but interval $[2,4]$ has all positive values.
No, $f(x)$ is never $0$ on $[2, 4]$. Function equals zero at $x = \pm 2$, but interval $[2,4]$ has all positive values.
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Verify IVT for $f(x) = x^3 + x$ on $[0, 1]$, $N = 0.5$.
Verify IVT for $f(x) = x^3 + x$ on $[0, 1]$, $N = 0.5$.
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Yes, $f(0) < 0.5 < f(1)$. $f(0) = 0 < 0.5 < 2 = f(1)$ confirms IVT applies with continuity.
Yes, $f(0) < 0.5 < f(1)$. $f(0) = 0 < 0.5 < 2 = f(1)$ confirms IVT applies with continuity.
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Does $f(x) = \text{cos}(x)$ satisfy IVT on $[0, \frac{3\text{π}}{2}]$, $N = 0$?
Does $f(x) = \text{cos}(x)$ satisfy IVT on $[0, \frac{3\text{π}}{2}]$, $N = 0$?
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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.
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.
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