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Tangent Line Approximations

There are a number of applications of derivatives. The linear approximation is one of them. The local linearity is the basic idea behind the concept of linear approximation. According to this approach, a tangent to a function near the point of tangency lies much closer to the function. Hence, it can be used to approximate the function. Thus, the linear approximation is based on the closeness of tangent to a curve of a function near a given point. Let us go ahead and learn the concept of linear approximation in detail.

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The estimated value of a function at a point is close to the value of linear function having its graph a tangent line. Therefore, the linear approximation of a function $f(x)$ near $x$ = $k$ is said to be the linear function whose graph is a tangent to the function $f(x)$ at given point $(k,\ f(k))$.
Tangent Line Approximations
For linear approximation formula, we need the equation of tangent line at point $(k,\ f(k))$. We know that the slope of tangent line at $(k,\ f(k))$ is $f'\ (k)$, so equation of tangent line is :

$y$ = $y_{0}\ +\ m(x\ -\ x_{0})$

$y$ = $f(k)\ +\ f'\ (k)(x\ -\ k)$

Thus, the equation of linear approximation of $f(x)$ near $x$ = $k$ is given by :
$f(x) \approx f(k) + f'(k)(x - k)$

Finding Tangent Line Approximation

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The following steps are to be following while finding linear approximation of a function.

Step 1 : Identify the given function. For example, in order to approximate $\sqrt{4.2}$, the function would be $f(x)$ = $\sqrt{x}$.

Step 2 : Determine a number that is close to the given number. Here, this number can be $4$ since it is quite close to the $4.2$.

Step 3 : Find the required point and slope of the tangent line at that point. i.e.

             $f(4)$ = $\sqrt{4}$ = $2$. So, the point is $(4,\ 2)$

             And the slope is

             $f'(x)$ = $\frac{1}{2 \sqrt{x}}$ or $f'(4)$ = $\frac{1}{2 \sqrt{4}}$ = $\frac{1}{4}$.

Step 4 : Compute the equation of tangent line at that number. i.e.

             $y$ = $f(k)\ +\ f'\ (k)(x\ -\ k)$

             $y$ = $2\ +$ $\frac{1}{4}$ $(x\ -\ 4)$ = $\frac{x}{4}$ $+\ 1$

Step 5 : By plugging the given value (to be approximated) in this tangent line, we estimate our answer. Hence, for $x$ = $4.2$, we get

             $y$ = $\frac{4.2}{4}$ $+\ 1$

             $y$ = $2.05$

Local linear Approximation

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Let us consider a function $f(x)$. Now, if x is close to point $k$, then the value of given function at x is very very close to the value of height of the tangent line to the function $f(x)$ at point $k$, which is given by :

$f(k)\ +\ f'\ (k )\ (x\ -\ k)$

Local linear Approximation

Hence, the function $f(k)\ +\ f' (k)\ (x - k)$ is known as the local linear approximation of the function f at point $k$. This is a very good approximation for $f(x)$ if $x$ is closer to $k$, since the closer the two points are, the better the approximation is.

Piecewise Linear Approximation

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A piecewise linear function is defined as a real-valued function whose graph is composed of sections of the straight line. This function is typically not continuous. It is a denoted piecewise on the graph, where each of the pieces is said to be an affine function.

The linear approximation of a piecewise linear function at a given point is thus known as the piecewise linear approximation of the function.


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The examples based on linear approximation are discussed below.

Example 1: Evaluate the linear approximation of the function $f(x)$ = $x^{3}$ at point $a$ = $1$. Also find the value of $(1.004)^{3}$.

Solution: $f(x)$ = $x^{3}$

$f(1)$ = $1$

$f'(x)$ = $3x^{2}$

$f'(1)$ = $3$

Formula is :

$f(x)$ = $f(a)\ +\ f'(a)(x\ -\ a)$

$f(x)$ = $f(1)\ +\ f'(1)(x\ -\ 1)$

$f(x)$ = $1\ +\ 3(x\ -\ 1)$

$f(x)$ = $3x\ -\ 2$

Which the linear approximation of given function at $x$ = $1$.

Plugging $1.004$, we get

$f(1.004)$ = $(1.004)^{3}$

= $3(1.004)\ -\ 2$ = $1.012$
Example 2: Using linear approximation method, approximate $\sqrt[3]{8.05}$.

Solution: The function would be $f(x)$ = $\sqrt[3]{x}$

Let's take $k$ = $8$

$f(8)$ = $\sqrt[3]{8}$ = $2$

$f'(x)$ = $\frac{1}{3}$ $x$$^{-\frac{2}{3}}$

$f'(8)$ = $\frac{1}{3}$ $8$$^{\frac{2}{3}}$ = $\frac{1}{12}$

Formula is :

$f(x)$ = $f(k)\ +\ f'(k)(x\ -\ k)$

$f(x)$ = $f(8)\ +\ f'(8)(x\ -\ 8)$

$f(x)$ = $2\ +$ $\frac{1}{12}$$(x\ -\ 8)$

$f(x)$ = $2\ +$ $\frac{x}{12}x$ $-$ $\frac{2}{3}$

$f(x)$ = $\frac{x}{12}$ $+$ $\frac{4}{3}$

Which the linear approximation of given function at $x$ = $8$.

Plugging $x$ = $8.05$, we get

$f(8.05)$ = $\frac{8.05}{12}$ $+$ $\frac{4}{3}$

= $0.671\ +\ 1.333$

= $2.004$

= $3(1.004)\ -\ 2$ = $1.012$
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