Find the taylor polynomial t1(x) for the function f(x)=ln(x) based at b=1.

Assuming [tex]T_1(x)[/tex] refers to the first degree Taylor polynomial, you need only find the value of [tex]f[/tex] and its derivative at [tex]x=1[/tex]. You have [tex]f(1)=\ln1=0[/tex] and [tex]f'(1)=\dfrac11=1[/tex].

The first degree Taylor approximation to [tex]f[/tex] centered at [tex]b=1[/tex] is then [tex]T_1(x)=f(1)+f'(1)(x-1)=x-1[/tex].

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keshavgandhi04 keshavgandhi04 · Answer 2 · 2021-12-04T11:53:52+01:00

The taylor polynomial for the function (f(x)=ln(x)) based at b=1 is (f(x) = x-1) and this can be determined by using the Taylor first degree approximation formula.

Given :

f(x) = ln(x)

The following steps can be used n order to determine the Taylor polynomial:

Step 1 - The Taylor approximation is given by:

[tex]\rm f(x) = f(b) + \dfrac{f'(b)}{1!}(x-b) + \dfrac{f''(b)}{2!}(x-b)^2+ \dfrac{f'''(b)}{3!}(x-b)^3+....[/tex]

Step 2 - Now, differentiate the given function f(x) = ln(x).

[tex]\rm f'(x) = \dfrac{1}{x}[/tex]

Step 3 - Now, substitute the value of (x = 1) in the above expression.

[tex]\rm f'(1) = \dfrac{1}{1}=1[/tex]

Step 3 - Now, the first degree Taylor approximation is given by:

[tex]\rm f(x) = f(b) + \dfrac{f'(b)}{1!}(x-b)[/tex]

Step 4 - Substitute the values of the known terms in the above expression.

[tex]\rm f(x) = f(1) + \dfrac{f'(1)}{1!}(x-1)[/tex]

Step 5 - Simplify the above expression.

f(x) = ln(1) + (x - 1)

f(x) = x - 1

So, the taylor polynomial for the function (f(x)=ln(x)) based at b=1 is (f(x) = x-1).

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https://brainly.com/question/2096984