How do you find the second-degree Taylor polynomial #T_2(x)# for the function #f(x) = sqrt(3+x^2)# at the number x=1?

1 Answer
Mar 13, 2017

#T_2(x)=2+1/2(x-1)+3/16(x-1)^2#

Explanation:

A complete Taylor polynomial for the function #f# centered around #x=c# is given by:

#T(x)=sum_(n=0)^oo(f^((n))(c))/(n!)(x-c)^n#

So the second degree Taylor polynomial will be the sum of the terms #n=0# through #n=2#, or:

#T_2(x)=(f^((0))(c))/(0!)(x-c)^0+(f^((1))(c))/(1!)(x-c)^1+(f^((2))(c))/(2!)(x-c)^2" "#

#c=1# and we can make a couple other simplifications:

#T_2(x)=f(1)+f'(1)(x-1)+(f''(1))/2(x-1)^2" "" "color(red)star#

We see that:

#f(1)=sqrt(3+1)=2#

And:

#f'(x)=1/2(3+x^2)^(-1/2)(2x)=x/sqrt(3+x^2)#

#f'(1)=1/sqrt(3+1)=1/2#

Finding #f''# using #f'(x)=x(3+x^2)^(-1/2)#:

#f''(x)=(3+x^2)^(-1/2)+x(-1/2(3+x^2)^(-3/2))(2x)#

#f''(x)=(3+x^2)^(-1/2)-x^2(3+x^2)^(-3/2)#

#f''(x)=3/(3+x^2)^(3/2)#

#f''(1)=3/(1+3)^(3/2)=3/8#

Returning to #color(red)star#:

#T_2(x)=2+1/2(x-1)+3/16(x-1)^2#