Connecting Differentiability and Continuity - AP Calculus BC
Card 1 of 30
Determine if $f(x) = \text{ln}(x)$ is differentiable at $x=0$.
Determine if $f(x) = \text{ln}(x)$ is differentiable at $x=0$.
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No, it is not differentiable at $x=0$. Natural logarithm is undefined at zero, so no derivative exists.
No, it is not differentiable at $x=0$. Natural logarithm is undefined at zero, so no derivative exists.
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What is the derivative of $f(x) = \text{ln}(x)$ for $x > 0$?
What is the derivative of $f(x) = \text{ln}(x)$ for $x > 0$?
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$f'(x) = \frac{1}{x}$. This is the standard derivative formula for natural logarithm.
$f'(x) = \frac{1}{x}$. This is the standard derivative formula for natural logarithm.
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Find the derivative of $f(x) = |x|$ at $x = 0$.
Find the derivative of $f(x) = |x|$ at $x = 0$.
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The derivative does not exist at $x = 0$. The absolute value function has a sharp corner at the origin.
The derivative does not exist at $x = 0$. The absolute value function has a sharp corner at the origin.
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What is the relationship between left-hand and right-hand derivatives for existence?
What is the relationship between left-hand and right-hand derivatives for existence?
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They must be equal for the derivative to exist. Left and right limits of the difference quotient must match.
They must be equal for the derivative to exist. Left and right limits of the difference quotient must match.
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Identify if differentiability implies continuity.
Identify if differentiability implies continuity.
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Differentiability implies continuity. If a function has a derivative, it must be continuous.
Differentiability implies continuity. If a function has a derivative, it must be continuous.
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What is the definition of differentiability at a point?
What is the definition of differentiability at a point?
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A function is differentiable at $x=a$ if $f'(a)$ exists. The derivative exists when the limit of the difference quotient exists.
A function is differentiable at $x=a$ if $f'(a)$ exists. The derivative exists when the limit of the difference quotient exists.
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Find $f'(x)$ for $f(x) = 3x^3 - 4x + 1$ at $x=1$.
Find $f'(x)$ for $f(x) = 3x^3 - 4x + 1$ at $x=1$.
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$f'(1) = 5$. Power rule gives $f'(x) = 9x^2 - 4$, so $f'(1) = 5$.
$f'(1) = 5$. Power rule gives $f'(x) = 9x^2 - 4$, so $f'(1) = 5$.
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What is the derivative of $f(x) = e^x$ at any point $x$?
What is the derivative of $f(x) = e^x$ at any point $x$?
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$f'(x) = e^x$. The exponential function is its own derivative everywhere.
$f'(x) = e^x$. The exponential function is its own derivative everywhere.
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State the rule for the existence of derivatives related to sharp corners.
State the rule for the existence of derivatives related to sharp corners.
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Derivatives do not exist at sharp corners. Sharp corners create different left and right-hand derivatives.
Derivatives do not exist at sharp corners. Sharp corners create different left and right-hand derivatives.
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Is $f(x) = x^2\text{sin}(\frac{1}{x})$ continuous at $x=0$?
Is $f(x) = x^2\text{sin}(\frac{1}{x})$ continuous at $x=0$?
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Yes, it is continuous at $x=0$. The squeeze theorem shows continuity despite oscillation.
Yes, it is continuous at $x=0$. The squeeze theorem shows continuity despite oscillation.
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Determine if $f(x) = x^4$ is differentiable at $x=0$.
Determine if $f(x) = x^4$ is differentiable at $x=0$.
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Yes, $f(x) = x^4$ is differentiable at $x=0$. Even powers create smooth curves differentiable at all points.
Yes, $f(x) = x^4$ is differentiable at $x=0$. Even powers create smooth curves differentiable at all points.
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Find if $f(x) = x^3$ is differentiable at $x=0$.
Find if $f(x) = x^3$ is differentiable at $x=0$.
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Yes, $f(x) = x^3$ is differentiable at $x=0$. Polynomial functions are differentiable everywhere in their domain.
Yes, $f(x) = x^3$ is differentiable at $x=0$. Polynomial functions are differentiable everywhere in their domain.
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Find if $f(x) = \text{sgn}(x)$ is continuous at $x=0$.
Find if $f(x) = \text{sgn}(x)$ is continuous at $x=0$.
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$f(x) = \text{sgn}(x)$ is not continuous at $x=0$. The sign function has a jump discontinuity at the origin.
$f(x) = \text{sgn}(x)$ is not continuous at $x=0$. The sign function has a jump discontinuity at the origin.
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Does $f(x) = \text{abs}(x)$ have a derivative at $x=0$?
Does $f(x) = \text{abs}(x)$ have a derivative at $x=0$?
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No, it does not have a derivative at $x=0$. Absolute value creates a sharp corner with undefined derivative.
No, it does not have a derivative at $x=0$. Absolute value creates a sharp corner with undefined derivative.
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State if $f(x) = [x]$ (floor function) is differentiable at $x=1$.
State if $f(x) = [x]$ (floor function) is differentiable at $x=1$.
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No, it is not differentiable at $x=1$. Floor functions have jump discontinuities at integer values.
No, it is not differentiable at $x=1$. Floor functions have jump discontinuities at integer values.
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Identify if continuity implies differentiability.
Identify if continuity implies differentiability.
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Continuity does not imply differentiability. Continuous functions can have corners where derivatives don't exist.
Continuity does not imply differentiability. Continuous functions can have corners where derivatives don't exist.
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Identify if $f(x) = \text{exp}(x)$ is differentiable at $x=1$.
Identify if $f(x) = \text{exp}(x)$ is differentiable at $x=1$.
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Yes, $f(x) = \text{exp}(x)$ is differentiable at $x=1$. Exponential functions are differentiable everywhere in their domain.
Yes, $f(x) = \text{exp}(x)$ is differentiable at $x=1$. Exponential functions are differentiable everywhere in their domain.
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Find if $f(x) = \frac{1}{x-1}$ is continuous at $x=1$.
Find if $f(x) = \frac{1}{x-1}$ is continuous at $x=1$.
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$f(x)$ is not continuous at $x=1$. Division by zero creates a vertical asymptote and discontinuity.
$f(x)$ is not continuous at $x=1$. Division by zero creates a vertical asymptote and discontinuity.
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State the rule for the existence of derivatives related to discontinuities.
State the rule for the existence of derivatives related to discontinuities.
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Derivatives do not exist at discontinuities. Differentiability requires continuity as a prerequisite.
Derivatives do not exist at discontinuities. Differentiability requires continuity as a prerequisite.
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Is $f(x) = \text{step}(x)$ differentiable at $x=2$?
Is $f(x) = \text{step}(x)$ differentiable at $x=2$?
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No, $f(x) = \text{step}(x)$ is not differentiable at $x=2$. Step functions have jump discontinuities at transition points.
No, $f(x) = \text{step}(x)$ is not differentiable at $x=2$. Step functions have jump discontinuities at transition points.
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State the rule for the existence of derivatives related to vertical tangents.
State the rule for the existence of derivatives related to vertical tangents.
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Derivatives do not exist at vertical tangents. Vertical tangents have infinite slope, making derivatives undefined.
Derivatives do not exist at vertical tangents. Vertical tangents have infinite slope, making derivatives undefined.
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Determine if $f(x) = \text{sin}(x)$ is differentiable at $x=0$.
Determine if $f(x) = \text{sin}(x)$ is differentiable at $x=0$.
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Yes, $f(x) = \text{sin}(x)$ is differentiable at $x=0$. Sine function is smooth and differentiable everywhere.
Yes, $f(x) = \text{sin}(x)$ is differentiable at $x=0$. Sine function is smooth and differentiable everywhere.
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Is $f(x) = x \times \text{sgn}(x)$ differentiable at $x=0$?
Is $f(x) = x \times \text{sgn}(x)$ differentiable at $x=0$?
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No, it is not differentiable at $x=0$. This equals $|x|$, which has a corner at the origin.
No, it is not differentiable at $x=0$. This equals $|x|$, which has a corner at the origin.
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For $f(x) = \tan(x)$, find $f'(x)$ at $x=\frac{\text{π}}{4}$.
For $f(x) = \tan(x)$, find $f'(x)$ at $x=\frac{\text{π}}{4}$.
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$f'(\frac{\text{π}}{4}) = 2$. Derivative of tangent is $\sec^2(x)$, which equals 2 at $\frac{\pi}{4}$.
$f'(\frac{\text{π}}{4}) = 2$. Derivative of tangent is $\sec^2(x)$, which equals 2 at $\frac{\pi}{4}$.
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Identify the differentiability of $f(x) = \frac{1}{2x}$ at $x=0$.
Identify the differentiability of $f(x) = \frac{1}{2x}$ at $x=0$.
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$f(x)$ is not differentiable at $x=0$. Division by zero makes the function undefined at the origin.
$f(x)$ is not differentiable at $x=0$. Division by zero makes the function undefined at the origin.
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Identify if continuity implies differentiability.
Identify if continuity implies differentiability.
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Continuity does not imply differentiability. Continuous functions can have corners where derivatives don't exist.
Continuity does not imply differentiability. Continuous functions can have corners where derivatives don't exist.
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Does $f(x) = \text{abs}(x)$ have a derivative at $x=0$?
Does $f(x) = \text{abs}(x)$ have a derivative at $x=0$?
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No, it does not have a derivative at $x=0$. Absolute value creates a sharp corner with undefined derivative.
No, it does not have a derivative at $x=0$. Absolute value creates a sharp corner with undefined derivative.
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What is the derivative of $f(x) = e^x$ at any point $x$?
What is the derivative of $f(x) = e^x$ at any point $x$?
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$f'(x) = e^x$. The exponential function is its own derivative everywhere.
$f'(x) = e^x$. The exponential function is its own derivative everywhere.
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Identify if differentiability implies continuity.
Identify if differentiability implies continuity.
Tap to reveal answer
Differentiability implies continuity. If a function has a derivative, it must be continuous.
Differentiability implies continuity. If a function has a derivative, it must be continuous.
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Find the derivative of $f(x) = |x|$ at $x = 0$.
Find the derivative of $f(x) = |x|$ at $x = 0$.
Tap to reveal answer
The derivative does not exist at $x = 0$. The absolute value function has a sharp corner at the origin.
The derivative does not exist at $x = 0$. The absolute value function has a sharp corner at the origin.
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