Finding Taylor Polynomial of Degree 4 for f(x)=sqrt(x) About a=4

[meganlz09]

I need to find the Taylor polynomial of degree 4 expanded about a=4 for the function f(x)=squareroot of (x)=x^(1/2)

This is what I've started with but I'm not sure how to proceed and if I even started correctly:
f'(x)(-1/2)x^(-1/2)=1/2sqrt(x)
f"(x)=(-1/4)x^(-3/2)=-1/4x^3/2
f"'(x)=(3/8)x^(-3)=3/8sqrt(x)
f""(x)=(-9/8)x^(-4)
and then i just plug 4 in for x

any explanation toward the correct answer would be great,thanks

 

[rootX]

it's

P(x) = f(x) + (x-a)*f'(a) + 1/2! * (x-a)^2 * f''(a) + ... +1/4! * (x-a)^4 * f''''(x)

so you need to plug in a and f(a),.. values

and
if f(x) = (x)^1/2
then f'(x) = 1/2(x)^-0.5 .. (your differentiation seems wrong)

 

[Billygoat]

Well now just plug into the taylor series...

f(x) = f(4) + f'(4)(x-4)+f''(4)(x-4)^2/2!+f'''(4)(x-4)^3/3!+...

Hope this helps...

 

[TheoMcCloskey]

However, you may want to check those derivatives again before proceeding, especially the last two.

 

[tron_2.0]

Well now just plug into the taylor series...

f(x) = f(4) + f'(4)(x-4)+f''(4)(x-4)^2/2!+f'''(4)(x-4)^3/3!+...

Hope this helps...

dont forget to divide by 0!, 1!, 2!, 3!,... accordingly