Local Linearity and Linearization - AP Calculus AB
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What is the approximation $L(x)$ when $f(x) = \text{tan}(x)$ at $x = 0$?
What is the approximation $L(x)$ when $f(x) = \text{tan}(x)$ at $x = 0$?
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$L(x) = x$. Since $\tan(0) = 0$ and $\tan'(0) = \sec^2(0) = 1$.
$L(x) = x$. Since $\tan(0) = 0$ and $\tan'(0) = \sec^2(0) = 1$.
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What is the approximation $L(x)$ when $f(x) = \text{tan}(x)$ at $x = 0$?
What is the approximation $L(x)$ when $f(x) = \text{tan}(x)$ at $x = 0$?
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$L(x) = x$. Since $\tan(0) = 0$ and $\tan'(0) = \sec^2(0) = 1$.
$L(x) = x$. Since $\tan(0) = 0$ and $\tan'(0) = \sec^2(0) = 1$.
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Identify the derivative required for linearization of $f(x)$ at $x = a$.
Identify the derivative required for linearization of $f(x)$ at $x = a$.
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$f'(a)$. The slope of the tangent line at the linearization point.
$f'(a)$. The slope of the tangent line at the linearization point.
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Which function value is used in the linear approximation $L(x) = f(a) + f'(a)(x - a)$?
Which function value is used in the linear approximation $L(x) = f(a) + f'(a)(x - a)$?
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$f(a)$. The y-intercept when the tangent line passes through $(a, f(a))$.
$f(a)$. The y-intercept when the tangent line passes through $(a, f(a))$.
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Determine the linearization of $f(x) = x^2$ at $x = 2$.
Determine the linearization of $f(x) = x^2$ at $x = 2$.
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$L(x) = 4 + 4(x - 2)$. Since $f(2) = 4$ and $f'(x) = 2x$, so $f'(2) = 4$.
$L(x) = 4 + 4(x - 2)$. Since $f(2) = 4$ and $f'(x) = 2x$, so $f'(2) = 4$.
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Identify the linear approximation for $f(x) = \text{exp}(x)$ at $x = 1$.
Identify the linear approximation for $f(x) = \text{exp}(x)$ at $x = 1$.
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$L(x) = e + e(x - 1)$. Since $e^1 = e$ and $(e^x)' = e^x$, so $f'(1) = e$.
$L(x) = e + e(x - 1)$. Since $e^1 = e$ and $(e^x)' = e^x$, so $f'(1) = e$.
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What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = 1$?
What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = 1$?
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$L(x) = x - 1$. Since $\ln(1) = 0$ and $(\ln x)' = \frac{1}{x}$, so $f'(1) = 1$.
$L(x) = x - 1$. Since $\ln(1) = 0$ and $(\ln x)' = \frac{1}{x}$, so $f'(1) = 1$.
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Which point is used as the center for a linear approximation?
Which point is used as the center for a linear approximation?
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The point $a$ where the approximation is centered. The base point where the tangent line touches the curve.
The point $a$ where the approximation is centered. The base point where the tangent line touches the curve.
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Identify the linear approximation of $f(x) = \text{cos}(x)$ at $x = 0$.
Identify the linear approximation of $f(x) = \text{cos}(x)$ at $x = 0$.
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$L(x) = 1$. Since $\cos(0) = 1$ and $\cos'(0) = -\sin(0) = 0$.
$L(x) = 1$. Since $\cos(0) = 1$ and $\cos'(0) = -\sin(0) = 0$.
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Which function approximates $f(x)$ near $x = a$ using local linearity?
Which function approximates $f(x)$ near $x = a$ using local linearity?
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The linearization $L(x)$ of $f(x)$. The best linear approximation near the given point.
The linearization $L(x)$ of $f(x)$. The best linear approximation near the given point.
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Find the linear approximation of $f(x) = \text{arcsin}(x)$ at $x = 0$.
Find the linear approximation of $f(x) = \text{arcsin}(x)$ at $x = 0$.
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$L(x) = x$. Since $\arcsin(0) = 0$ and $(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}$.
$L(x) = x$. Since $\arcsin(0) = 0$ and $(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}$.
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Calculate the derivative needed for linearization of $f(x) = \frac{1}{x}$ at $x = 1$.
Calculate the derivative needed for linearization of $f(x) = \frac{1}{x}$ at $x = 1$.
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$f'(x) = -\frac{1}{x^2}$, so $f'(1) = -1$. Using the power rule and evaluating at $x = 1$.
$f'(x) = -\frac{1}{x^2}$, so $f'(1) = -1$. Using the power rule and evaluating at $x = 1$.
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What is the linear approximation of $f(x) = \text{arctan}(x)$ at $x = 0$?
What is the linear approximation of $f(x) = \text{arctan}(x)$ at $x = 0$?
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$L(x) = x$. Since $\arctan(0) = 0$ and $(\arctan x)' = \frac{1}{1+x^2}$.
$L(x) = x$. Since $\arctan(0) = 0$ and $(\arctan x)' = \frac{1}{1+x^2}$.
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What is the linear approximation of $f(x) = e^x$ at $x = 0$?
What is the linear approximation of $f(x) = e^x$ at $x = 0$?
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$L(x) = 1 + x$. Since $e^0 = 1$ and $(e^x)' = e^x$, so $f'(0) = 1$.
$L(x) = 1 + x$. Since $e^0 = 1$ and $(e^x)' = e^x$, so $f'(0) = 1$.
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State the condition under which a function is locally linear at a point.
State the condition under which a function is locally linear at a point.
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The derivative $f'(x)$ exists and is continuous near the point. Ensures the function behaves like a line near that point.
The derivative $f'(x)$ exists and is continuous near the point. Ensures the function behaves like a line near that point.
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Find the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$.
Find the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$.
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$L(x) = x$. Since $\sin(0) = 0$ and $\sin'(0) = \cos(0) = 1$.
$L(x) = x$. Since $\sin(0) = 0$ and $\sin'(0) = \cos(0) = 1$.
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Calculate the derivative needed for linearization of $f(x) = \frac{1}{x}$ at $x = 1$.
Calculate the derivative needed for linearization of $f(x) = \frac{1}{x}$ at $x = 1$.
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$f'(x) = -\frac{1}{x^2}$, so $f'(1) = -1$. Using the power rule and evaluating at $x = 1$.
$f'(x) = -\frac{1}{x^2}$, so $f'(1) = -1$. Using the power rule and evaluating at $x = 1$.
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What is the linear approximation of $f(x) = x^4$ at $x = 1$?
What is the linear approximation of $f(x) = x^4$ at $x = 1$?
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$L(x) = 1 + 4(x - 1)$. Since $f(1) = 1$ and $f'(x) = 4x^3$, so $f'(1) = 4$.
$L(x) = 1 + 4(x - 1)$. Since $f(1) = 1$ and $f'(x) = 4x^3$, so $f'(1) = 4$.
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What is the linear approximation of $f(x) = \text{arctan}(x)$ at $x = 0$?
What is the linear approximation of $f(x) = \text{arctan}(x)$ at $x = 0$?
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$L(x) = x$. Since $\arctan(0) = 0$ and $(\arctan x)' = \frac{1}{1+x^2}$.
$L(x) = x$. Since $\arctan(0) = 0$ and $(\arctan x)' = \frac{1}{1+x^2}$.
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State the condition under which a function is locally linear at a point.
State the condition under which a function is locally linear at a point.
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The derivative $f'(x)$ exists and is continuous near the point. Ensures the function behaves like a line near that point.
The derivative $f'(x)$ exists and is continuous near the point. Ensures the function behaves like a line near that point.
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Find the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$.
Find the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$.
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$L(x) = x$. Since $\sin(0) = 0$ and $\sin'(0) = \cos(0) = 1$.
$L(x) = x$. Since $\sin(0) = 0$ and $\sin'(0) = \cos(0) = 1$.
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Identify the derivative required for linearization of $f(x)$ at $x = a$.
Identify the derivative required for linearization of $f(x)$ at $x = a$.
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$f'(a)$. The slope of the tangent line at the linearization point.
$f'(a)$. The slope of the tangent line at the linearization point.
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What is the formula for the linear approximation of a function at point $a$?
What is the formula for the linear approximation of a function at point $a$?
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$L(x) = f(a) + f'(a)(x - a)$. This is the tangent line equation at point $a$.
$L(x) = f(a) + f'(a)(x - a)$. This is the tangent line equation at point $a$.
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What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = 1$?
What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = 1$?
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$L(x) = x - 1$. Since $\ln(1) = 0$ and $(\ln x)' = \frac{1}{x}$, so $f'(1) = 1$.
$L(x) = x - 1$. Since $\ln(1) = 0$ and $(\ln x)' = \frac{1}{x}$, so $f'(1) = 1$.
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Which point is used as the center for a linear approximation?
Which point is used as the center for a linear approximation?
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The point $a$ where the approximation is centered. The base point where the tangent line touches the curve.
The point $a$ where the approximation is centered. The base point where the tangent line touches the curve.
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Find the linear approximation of $f(x) = \text{arcsin}(x)$ at $x = 0$.
Find the linear approximation of $f(x) = \text{arcsin}(x)$ at $x = 0$.
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$L(x) = x$. Since $\arcsin(0) = 0$ and $(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}$.
$L(x) = x$. Since $\arcsin(0) = 0$ and $(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}$.
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What is the linear approximation for $f(x) = \text{sqrt}(x)$ at $x = 4$?
What is the linear approximation for $f(x) = \text{sqrt}(x)$ at $x = 4$?
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$L(x) = 2 + \frac{1}{4}(x - 4)$. Since $\sqrt{4} = 2$ and $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$.
$L(x) = 2 + \frac{1}{4}(x - 4)$. Since $\sqrt{4} = 2$ and $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$.
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State the linear approximation for $f(x) = \frac{1}{x}$ at $x = 2$.
State the linear approximation for $f(x) = \frac{1}{x}$ at $x = 2$.
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$L(x) = \frac{1}{2} - \frac{1}{4}(x - 2)$. Since $f(2) = \frac{1}{2}$ and $f'(2) = -\frac{1}{4}$.
$L(x) = \frac{1}{2} - \frac{1}{4}(x - 2)$. Since $f(2) = \frac{1}{2}$ and $f'(2) = -\frac{1}{4}$.
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Identify the linear approximation for $f(x) = \text{exp}(x)$ at $x = 1$.
Identify the linear approximation for $f(x) = \text{exp}(x)$ at $x = 1$.
Tap to reveal answer
$L(x) = e + e(x - 1)$. Since $e^1 = e$ and $(e^x)' = e^x$, so $f'(1) = e$.
$L(x) = e + e(x - 1)$. Since $e^1 = e$ and $(e^x)' = e^x$, so $f'(1) = e$.
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What is the linear approximation of $f(x) = x^5$ at $x = 1$?
What is the linear approximation of $f(x) = x^5$ at $x = 1$?
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$L(x) = 1 + 5(x - 1)$. Since $f(1) = 1$ and $f'(x) = 5x^4$, so $f'(1) = 5$.
$L(x) = 1 + 5(x - 1)$. Since $f(1) = 1$ and $f'(x) = 5x^4$, so $f'(1) = 5$.
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