Local Linearity and Linearization - AP Calculus BC
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What is the linear approximation of $f(x) = \frac{1}{x}$ at $x = 3$?
What is the linear approximation of $f(x) = \frac{1}{x}$ at $x = 3$?
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$L(x) = \frac{1}{3} - \frac{1}{9}(x-3)$. $f(3) = \frac{1}{3}$, $f'(3) = -\frac{1}{9}$, standard linearization formula.
$L(x) = \frac{1}{3} - \frac{1}{9}(x-3)$. $f(3) = \frac{1}{3}$, $f'(3) = -\frac{1}{9}$, standard linearization formula.
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Find the linear approximation of $f(x) = x^2$ at $x = 0$ for $x = 0.1$.
Find the linear approximation of $f(x) = x^2$ at $x = 0$ for $x = 0.1$.
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$L(x) = 0$. $f(0) = 0$, $f'(0) = 0$, so linearization gives zero.
$L(x) = 0$. $f(0) = 0$, $f'(0) = 0$, so linearization gives zero.
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What is $L(x)$ in the linear approximation formula?
What is $L(x)$ in the linear approximation formula?
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The linear approximation of $f(x)$ near $x = a$. The linearization function that approximates $f(x)$.
The linear approximation of $f(x)$ near $x = a$. The linearization function that approximates $f(x)$.
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Evaluate $L(x)$ for $f(x) = \text{sqrt}(x)$ at $x = 4$ using $x = 4.1$.
Evaluate $L(x)$ for $f(x) = \text{sqrt}(x)$ at $x = 4$ using $x = 4.1$.
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$L(x) = 2 + \frac{0.1}{4} = 2.025$. $f(4) = 2$, $f'(4) = \frac{1}{4}$, so $L(4.1) = 2 + \frac{1}{4}(0.1)$.
$L(x) = 2 + \frac{0.1}{4} = 2.025$. $f(4) = 2$, $f'(4) = \frac{1}{4}$, so $L(4.1) = 2 + \frac{1}{4}(0.1)$.
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Find the linear approximation of $f(x) = \frac{1}{x}$ at $x = 1$ for $x = 0.9$.
Find the linear approximation of $f(x) = \frac{1}{x}$ at $x = 1$ for $x = 0.9$.
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$L(x) = 1 + 0.1 = 1.1$. $f(1) = 1$, $f'(1) = -1$, so $L(0.9) = 1 + (-1)(-0.1)$.
$L(x) = 1 + 0.1 = 1.1$. $f(1) = 1$, $f'(1) = -1$, so $L(0.9) = 1 + (-1)(-0.1)$.
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What is the linear approximation of $f(x) = e^x$ at $x = 1$?
What is the linear approximation of $f(x) = e^x$ at $x = 1$?
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$L(x) = e + e(x-1)$. $f(1) = e$, $f'(1) = e$, standard exponential linearization.
$L(x) = e + e(x-1)$. $f(1) = e$, $f'(1) = e$, standard exponential linearization.
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Calculate the linear approximation of $f(x) = x^2$ at $x=1$ for $x=1.1$.
Calculate the linear approximation of $f(x) = x^2$ at $x=1$ for $x=1.1$.
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$1 + 2(0.1) = 1.2$. $f(1) = 1$, $f'(1) = 2$, so $L(1.1) = 1 + 2(0.1)$.
$1 + 2(0.1) = 1.2$. $f(1) = 1$, $f'(1) = 2$, so $L(1.1) = 1 + 2(0.1)$.
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Identify the derivative in the linearization formula $L(x) = f(a) + f'(a)(x-a)$.
Identify the derivative in the linearization formula $L(x) = f(a) + f'(a)(x-a)$.
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$f'(a)$. This term represents the slope of the tangent line.
$f'(a)$. This term represents the slope of the tangent line.
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Find the linear approximation of $f(x) = \text{tan}(x)$ at $x = 0$ for $x = 0.05$.
Find the linear approximation of $f(x) = \text{tan}(x)$ at $x = 0$ for $x = 0.05$.
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$L(x) = 0.05$. $f(0) = 0$, $f'(0) = 1$, so $L(0.05) = 0 + 1(0.05)$.
$L(x) = 0.05$. $f(0) = 0$, $f'(0) = 1$, so $L(0.05) = 0 + 1(0.05)$.
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What is the linear approximation of $f(x) = \text{exp}(x)$ at $x = 1$?
What is the linear approximation of $f(x) = \text{exp}(x)$ at $x = 1$?
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$L(x) = e + e(x-1)$. $f(1) = e$, $f'(1) = e$, so linearization has slope $e$.
$L(x) = e + e(x-1)$. $f(1) = e$, $f'(1) = e$, so linearization has slope $e$.
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Evaluate $L(x)$ for $f(x) = \frac{1}{x}$ at $x = 2$ using $x = 2.1$.
Evaluate $L(x)$ for $f(x) = \frac{1}{x}$ at $x = 2$ using $x = 2.1$.
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$L(x) = \frac{1}{2} - \frac{0.1}{4} = 0.475$. $f(2) = 0.5$, $f'(2) = -0.25$, so $L(2.1) = 0.5 - 0.25(0.1)$.
$L(x) = \frac{1}{2} - \frac{0.1}{4} = 0.475$. $f(2) = 0.5$, $f'(2) = -0.25$, so $L(2.1) = 0.5 - 0.25(0.1)$.
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Approximate $\text{ln}(1.1)$ using linearization at $x = 1$ for $f(x) = \text{ln}(x)$.
Approximate $\text{ln}(1.1)$ using linearization at $x = 1$ for $f(x) = \text{ln}(x)$.
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$L(x) = 0 + 0.1 = 0.1$. $f(1) = 0$, $f'(1) = 1$, so $L(1.1) = 0 + 1(0.1)$.
$L(x) = 0 + 0.1 = 0.1$. $f(1) = 0$, $f'(1) = 1$, so $L(1.1) = 0 + 1(0.1)$.
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Calculate the linear approximation of $f(x) = x^3$ at $x = 0$ for $x = 0.1$.
Calculate the linear approximation of $f(x) = x^3$ at $x = 0$ for $x = 0.1$.
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$L(x) = 0$. $f(0) = 0$, $f'(0) = 0$, so linearization gives zero.
$L(x) = 0$. $f(0) = 0$, $f'(0) = 0$, so linearization gives zero.
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Determine the linear approximation of $f(x) = x^4$ at $x = 1$ for $x = 1.1$.
Determine the linear approximation of $f(x) = x^4$ at $x = 1$ for $x = 1.1$.
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$L(x) = 1 + 0.4 = 1.4$. $f(1) = 1$, $f'(1) = 4$, so $L(1.1) = 1 + 4(0.1)$.
$L(x) = 1 + 0.4 = 1.4$. $f(1) = 1$, $f'(1) = 4$, so $L(1.1) = 1 + 4(0.1)$.
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Approximate $e^{0.2}$ using linearization at $x = 0$ for $f(x) = e^x$.
Approximate $e^{0.2}$ using linearization at $x = 0$ for $f(x) = e^x$.
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$L(x) = 1 + 0.2 = 1.2$. $f(0) = 1$, $f'(0) = 1$, so $L(0.2) = 1 + 1(0.2)$.
$L(x) = 1 + 0.2 = 1.2$. $f(0) = 1$, $f'(0) = 1$, so $L(0.2) = 1 + 1(0.2)$.
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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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$L(x) = x - 1$. $f(1) = 0$, $f'(1) = 1$, giving the simple linear form.
$L(x) = x - 1$. $f(1) = 0$, $f'(1) = 1$, giving the simple linear form.
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Approximate $\text{sin}(0.1)$ using linearization at $x = 0$ for $f(x) = \text{sin}(x)$.
Approximate $\text{sin}(0.1)$ using linearization at $x = 0$ for $f(x) = \text{sin}(x)$.
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$L(x) = 0.1$. $f(0) = 0$, $f'(0) = 1$, so $L(0.1) = 0 + 1(0.1)$.
$L(x) = 0.1$. $f(0) = 0$, $f'(0) = 1$, so $L(0.1) = 0 + 1(0.1)$.
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Determine the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$ for $x = e+0.1$.
Determine the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$ for $x = e+0.1$.
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$L(x) = 1 + 0.1/e$. $f(e) = 1$, $f'(e) = \frac{1}{e}$, so slope is $\frac{1}{e}$.
$L(x) = 1 + 0.1/e$. $f(e) = 1$, $f'(e) = \frac{1}{e}$, so slope is $\frac{1}{e}$.
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What is the linear approximation of $f(x) = \text{tan}(x)$ at $x = 0$?
What is the linear approximation of $f(x) = \text{tan}(x)$ at $x = 0$?
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For $x$ near $0$, $L(x) = x$. $f(0) = 0$, $f'(0) = \sec^2(0) = 1$, so linearization is $x$.
For $x$ near $0$, $L(x) = x$. $f(0) = 0$, $f'(0) = \sec^2(0) = 1$, so linearization is $x$.
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What is the role of the derivative in linear approximation?
What is the role of the derivative in linear approximation?
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It gives the slope of the tangent line. The derivative provides the rate of change for the approximation.
It gives the slope of the tangent line. The derivative provides the rate of change for the approximation.
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Determine the linear approximation of $f(x) = x^3$ at $x = 1$ for $x = 1.05$.
Determine the linear approximation of $f(x) = x^3$ at $x = 1$ for $x = 1.05$.
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$L(x) = 1 + 3(0.05) = 1.15$. $f(1) = 1$, $f'(1) = 3$, so $L(1.05) = 1 + 3(0.05)$.
$L(x) = 1 + 3(0.05) = 1.15$. $f(1) = 1$, $f'(1) = 3$, so $L(1.05) = 1 + 3(0.05)$.
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Find $L(x)$ for $f(x) = \text{ln}(x)$ at $x = 1$ when $x = 1.1$.
Find $L(x)$ for $f(x) = \text{ln}(x)$ at $x = 1$ when $x = 1.1$.
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$L(x) = 0 + \frac{1}{1}(0.1) = 0.1$. $f(1) = 0$, $f'(1) = 1$, so $L(1.1) = 0 + 1(0.1)$.
$L(x) = 0 + \frac{1}{1}(0.1) = 0.1$. $f(1) = 0$, $f'(1) = 1$, so $L(1.1) = 0 + 1(0.1)$.
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Approximate $e^{0.1}$ using linearization at $x = 0$ for $f(x) = e^x$.
Approximate $e^{0.1}$ using linearization at $x = 0$ for $f(x) = e^x$.
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$L(x) = 1 + 0.1 = 1.1$. $f(0) = 1$, $f'(0) = 1$, so $L(0.1) = 1 + 1(0.1)$.
$L(x) = 1 + 0.1 = 1.1$. $f(0) = 1$, $f'(0) = 1$, so $L(0.1) = 1 + 1(0.1)$.
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What does $f'(a)(x-a)$ represent in the linearization formula?
What does $f'(a)(x-a)$ represent in the linearization formula?
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The change in the linear approximation. This represents how much the function changes linearly.
The change in the linear approximation. This represents how much the function changes linearly.
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Which function value does $f(a)$ represent in the linear approximation formula?
Which function value does $f(a)$ represent in the linear approximation formula?
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The value of the function at $x = a$. This is the y-coordinate where the tangent line touches the curve.
The value of the function at $x = a$. This is the y-coordinate where the tangent line touches the curve.
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What is the geometric interpretation of a linearization at $x = a$?
What is the geometric interpretation of a linearization at $x = a$?
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It's the tangent line to the function at $x = a$. The linearization creates the tangent line at that specific point.
It's the tangent line to the function at $x = a$. The linearization creates the tangent line at that specific point.
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Define local linearity in the context of a differentiable function.
Define local linearity in the context of a differentiable function.
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Local linearity means the function appears linear near a point. The function's graph looks like a straight line when zoomed in close.
Local linearity means the function appears linear near a point. The function's graph looks like a straight line when zoomed in close.
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What is the approximated change in $f(x)$ for $f(x) = x^3$ at $x = 1$?
What is the approximated change in $f(x)$ for $f(x) = x^3$ at $x = 1$?
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$3(x-1)$. For $f(x) = x^3$ at $x=1$, $f'(1) = 3$ gives the linear change.
$3(x-1)$. For $f(x) = x^3$ at $x=1$, $f'(1) = 3$ gives the linear change.
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What is the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$?
What is the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$?
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For $x$ near $0$, $L(x) = x$. $f(0) = 0$, $f'(0) = 1$, so linearization is just $x$.
For $x$ near $0$, $L(x) = x$. $f(0) = 0$, $f'(0) = 1$, so linearization is just $x$.
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For which type of function can local linearity be used to approximate values?
For which type of function can local linearity be used to approximate values?
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Differentiable functions. Must have a derivative to calculate the tangent line slope.
Differentiable functions. Must have a derivative to calculate the tangent line slope.
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