Question

find the local linear approximation of f(x) = e^2x at x = 0.

Answer 1

A local linear approximation can be found by taking the first two terms of the Taylor series of the function.

For f(x) = e^2x at x = 0,

f(x) ≈ f(0) + f'(0)(x)
= 1 + 2x

Answer 2

The linear approximation of $f(x) = e^{2x$ at x = 0 is y = 1 - 2x.

What is linear approximation?

Linear approximation exists as a way we can utilize to correspond to the value of a function at a particular point.

The chosen value of $x=x_{0}$, the function contains the value $f(x_{0})$

The tangent line through $(x_{0},f(x_{0}))$ contains a slope $f'(x_{0})$

The linear equation can be thought of as an equation for the tangent line or as a linear function giving an approximation of f.

The point-slope form of the equation of the tangent line (the line through $(x_{0},f(x_{0}))$ with slope $f'(x_{0})$ is:

$y-y_{0} =f'(x)(x-x_{0})$

The linear approximation for f at $x_{0}$ is

$y=y_{0}+f'(x)(x-x_{0} )$

For $f(x)=e^{2x$ at $x_{0} =0$we get

f(0) = 1 and $f'(x)=2e^{2x}$ so $m = f'(0) = 2.$

The linear approximation is

y = 1− 2(x − 0)

y = 1 - 2x.

Therefore, the linear approximation of $f(x) = e^{2x$ at x = 0 is y = 1 - 2x.

To learn more about linear approximation refer to:

https://brainly.com/question/26849420

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