# Tangent Line Approximation

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Tangent line approximation : Consider any function that is differentiable at x = c so the equation of tangent line at (c,f(c)) is given by

This is called tangent line approximation or linear approximation of f at c.

According to the above graph y=f(x) is a curve and 'c' is any point on the curve. y= L(x) is as an approximation to the function, f(x) at x = c. In the above case we call tangent line as the linear approximation to the function at x = c. **y = f(c) + f '(c)(x - c )**This is called tangent line approximation or linear approximation of f at c.

## Examples on Tangent Line Approximation

**Example 1 :**Determine the linear approximation for f(x) = $\sqrt[3]{x}$ at x = 8. Also find the linear approximation of $\sqrt[3]{8.05}$

**Solution :**f(x) = $\sqrt[3]{x}$

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

f(8) = 2

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

c= 8 f '(c)|(c=8) = $\frac{1}{3}(8^{-2/3})$

= $\frac{1}{3}((2^{3})^{-2/3})$

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

f '(8) = $\frac{1}{3}*\frac{1}{4}$

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

Use a linear approximation formula,

**y = f(c) + f '(c)(x - c )**

c= 8 y = f(8) + f '(8)(x - 8 )

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

f(x) = $\frac{1}{12}x + \frac{4}{3}$

To find linear approximation of $\sqrt[3]{8.05}$ , plug in x = 8.05 in the above equation

f(25) = $\frac{1}{12}(8.05) + \frac{4}{3}$

$\sqrt[3]{25}$ = 2.004166

**Example 2 :**Find the linearization of f(x) = sin(x) at x= $\frac{\pi}{2}$

**Solution :** f(x) = sin(x)

f($\frac{\pi}{2}) = sin(\frac{\pi}{2}$)

f($\frac{\pi}{2}$) = 1

f(x) = sin(x)

f'(x) = cos(x)

f'($\frac{\pi}{2}) = cos(\frac{\pi}{2}$)

f '($\frac{\pi}{2}$) = 0

Use a linear approximation formula,

** y = f(c) + f '(c)(x - c ) **

c = $\frac{\pi}{2}$

f(x) =$ f(\frac{\pi}{2}) + f '(\frac{\pi}{2})(x - \frac{\pi}{2} ) $

f(x) = 1

So tangent line at x= $\frac{\pi}{2}$ is y=1

**12th grade math**

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