Linear Approximation

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In this lesson we will use the tangent line to approximate the value of a function near the point of tangency.

Given a function $y=f(x)$ , the equation of the tangent line at the point where $x = x_0$ is given by $y-f(x_0) = f'(x_0)(x - x_0)$ or $y = f(x_0) + f'(x_0)(x - x_0).$ The main idea of this section is that if we let $L(x) = f(x_0) + f'(x_0)(x - x_0)$ then $f(x_0) = L(x_0)$ and $f(x) \approx L(x)$ for values of $x$ close to $x_0$ . The function $L(x)$ is called the linearization of $f(x)$ at $x = x_0$ . The advantage of working with $L(x)$ is that values of a linear function are usually easy to compute. In a typical linear approximation problem, we are trying to approximate $f(x_1)$ and we need to choose $x_0$ and create $L(x)$ . Once we have accomplished this, our solution is $f(x_1) \approx L(x_1).$ There are two keys to choosing $x_0$ . First $x_0$ should be close to $x_1$ and second, we must be able to compute the exact value of $f(x_0)$ .

We can approximate $\sqrt {18}$ using linear approximation as follows. Let $f(x) = \sqrt x$ and since 16 is near 18 and $\sqrt {16} = 4$ , we let $x_0 = 16$ . To create $L(x)$ we also need to compute $f'(16)$ . Since $f'(x) = \frac {d}{dx} \sqrt x = \frac {d}{dx}(x^{1/2})= \frac {1}{2}x^{-1/2} = \frac {1}{2\sqrt x},$ we have $f'(16) = \frac {1}{2\sqrt {16}} = \frac {1}{8}.$

Next,

$\begin{align*} L(x) &= f(x_0) + f'(x_0)(x - x_0) \\ &= f(16) + f'(16)(x - 16) \\ &= 4 + \frac{1}{8}(x - 16) \end{align*}$ and finally, $\sqrt {18} \approx L(18) = 4 + \frac {1}{8}(18 - 16)$ $= 4 + \frac {2}{8} = \frac {17}{4} = 4.25.$

We can approximate $\sqrt {99}$ using linear approximation as follows. Let $f(x) = \sqrt x$ and since 100 is near 99 and $\sqrt {100} = 10$ , we let $x_0 = 100$ . To create $L(x)$ we also need to compute $f'(100)$ . Since $f'(x) = \frac {d}{dx} \sqrt x = \frac {d}{dx}(x^{1/2})= \frac {1}{2}x^{-1/2} = \frac {1}{2\sqrt x},$ we have $f'(100) = \frac {1}{2\sqrt {100}} = \frac {1}{20}.$

Next,

$\begin{align*} L(x) &= f(x_0) + f'(x_0)(x - x_0) \\ &= f(100) + f'(100)(x - 100) \\ &= 10 + \frac{1}{20}(x - 100) \end{align*}$ and finally, $\sqrt {99} \approx L(99) = 10 + \frac {1}{20}(99 - 100)$ $= 10 - \frac {1}{20} = \frac {199}{20} = 9.95.$

Find the linearization of $f(x) = \sqrt x$ at $x = 4$ and use it to approximate $\sqrt 5$ .

The linearization is $L(x) = \answer {2 + .25 (x-4)}$ .

The approximation is $\sqrt 5 \approx \answer {9/4}$ .

Find the linearization of $f(x) = \sqrt x$ at $x = 34$ and use it to approximate $\sqrt {34}$ .

The linearization is $L(x) = \answer {6 + (1/12) (x-36)}$ .

The approximation is $\sqrt {34} \approx \answer {35/6)}$ .

We can approximate $\sqrt [3]{10}$ using linear approximation as follows. Let $f(x) = \sqrt [3] x$ and since 10 is near 8 and $\sqrt [3]{8} = 2$ , we let $x_0 = 8$ . To create $L(x)$ we also need to compute $f'(8)$ . Since $f'(x) = \frac {d}{dx} \sqrt [3] x = \frac {d}{dx}(x^{1/3})= \frac {1}{3}x^{-2/3} = \frac {1}{3\sqrt [3] {x^2}},$ we have $f'(8) = \frac {1}{3\sqrt [3]{8^2}} =\frac {1}{3\sqrt [3]{64}}= \frac {1}{12}.$

Next,

$\begin{align*} L(x) &= f(x_0) + f'(x_0)(x - x_0) \\ &= f(8) + f'(8)(x - 8) \\ &= 2 + \frac{1}{12}(x - 8) \end{align*}$ and finally, $\sqrt [3]{10} \approx L(10) = 2 + \frac {1}{12}(10 - 8)$ $= 2 + \frac {2}{12} = \frac {13}{6} = 2.1{\overline 6}.$

Find the linearization of $f(x) = \sqrt [3] x$ at $x = 27$ and use it to approximate $\sqrt [3]{25}$ .

The linearization is $L(x) = \answer {3 + (1/27 (x-27)}$ .

The approximation is $\sqrt [3]{25} \approx \answer {79/27}$ .

We can approximate $e^{-0.1}$ using linear approximation as follows. Let $f(x) = e^x$ and since 0 is near $-0.1$ and $e^0 = 1$ , we let $x_0 = 0$ . To create $L(x)$ we also need to compute $f'(0)$ . Since $f'(x) = \frac {d}{dx} e^x = e^x,$ we have $f'(0) = e^0 = 1.$

Next,

$\begin{align*} L(x) &= f(x_0) + f'(x_0)(x - x_0) \\ &= f(0) + f'(0)(x - 0) \\ &= 1 + 1(x - 0)\\ & = 1 + x \end{align*}$ and finally, $e^{-0.1} \approx L(-0.1) = 1 - 0.1 = 0.9.$

Find the linearization of $f(x) = e^x$ at $x = 0$ and use it to approximate $e^{0.2}$ .

The linearization is $L(x) = \answer {1 + x}$ .

The approximation is $e^{0.2} \approx \answer {1.2}$ .

We can approximate $\ln (1.1)$ using linear approximation as follows. Let $f(x) = \ln x$ and since 1 is near 1.1 and $\ln (1) = 0$ , we let $x_0 = 1$ . To create $L(x)$ we also need to compute $f'(1)$ . Since $f'(x) = \frac {d}{dx} \ln x = \frac {1}{x},$ we have $f'(1) = \frac {1}{1} = 1.$

Next,

$\begin{align*} L(x) &= f(x_0) + f'(x_0)(x - x_0) \\ &= f(1) + f'(1)(x - 1) \\ &= 0 + 1(x - 1)\\ &=x-1 \end{align*}$ and finally, $\ln (1.1) \approx L(1.1) = 1.1 - 1 = 0.1.$

Find the linearization of $f(x) = \ln x$ at $x = 1$ and use it to approximate $\ln (0.8)$ .

The linearization is $L(x) = \answer {x-1}$ .

The approximation is $\ln (0.8) \approx \answer {-0.2}$ .

We can approximate $1.9^3$ using linear approximation as follows. Let $f(x) = x^3$ and since 2 is near 1.9 and $2^3 = 8$ , we let $x_0 = 2$ . To create $L(x)$ we also need to compute $f'(2)$ . Since $f'(x) = \frac {d}{dx} (x^3) = 3x^2,$ we have $f'(2) = 3(2^2) = 12.$

Next,

$\begin{align*} L(x) &= f(x_0) + f'(x_0)(x - x_0) \\ &= f(2) + f'(2)(x - 2) \\ &= 8 + 12(x - 2) \end{align*}$ and finally, $1.9^3 \approx L(2) = 8 + 12(1.9 - 2)$ $= 8 - 12(.1) = 8-(1.2) = 6.8.$

Find the linearization of $f(x) = x^4$ at $x = 1$ and use it to approximate $1.1^4$ .

The linearization is $L(x) = \answer {1 + 4 (x-1)}$ .

The approximation is $1.1^4 \approx \answer {1.4}$ .

We can approximate $\sin (1^{\circ })$ using linear approximation with the following twist. In the formula $\frac {d}{dx} \sin (x) = \cos (x)$ it is understood that the angle is measured in radians. Therefore, in order to use our linear approximation formula we need to restate our problem in radians as: $\text {Approximate} \ \ \sin \big (\frac {\pi }{180}\big ).$

We let $f(x) = \sin x$ and since 0 is near $\frac {\pi }{180}$ and $\sin (0) = 0$ , we let $x_0 = 0$ . To create $L(x)$ we also need to compute $f'(0)$ . Since $f'(x) = \frac {d}{dx} \sin (x) = \cos (x),$ we have $f'(0) = \cos (0) = 1.$

Next,

$\begin{align*} L(x) &= f(x_0) + f'(x_0)(x - x_0) \\ &= f(0) + f'(0)(x - 0) \\ &= 0 + 1(x - 0)\\ &= x \end{align*}$ and finally, $\sin (1^{\circ }) = \sin \big (\frac {\pi }{180}\big ) \approx L\big (\frac {\pi }{180}\big ) = \frac {\pi }{180}.$

Here is a detailed, lecture style video on linear approximation: