What is the quadratic taylor polynomial about 1 for the function f(x)sqrt x+8?

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1 Answer
Mar 10, 2018

#f(x)=-1/216x^2+19/108x+611/216#

Explanation:

#"using the "color(blue)"Taylor series"#

#•color(white)(x)f(x)=sum_(n=0)^oo(f^n(a))/(n!)(x-a)^n#

#color(white)(xxxxxx)=f(a)+(f'(a))/(1!)(x-a)+(f''(a))/(2!)(x-a)^2+...#

#"For a quadratic approximation we require to evaluate up"#
#"to and including the second degree term"#

#"here "a=1#

#rArrf(a)=f(1)=sqrt9=3#

#"differentiate using the "color(blue)"chain rule"#

#f(x)=sqrt(x+8)=(x+8)^(1/2)#

#rArrf'(x)=1/2(x+8)^(-1/2)#

#rArrf''(x)=-1/4(x+8)^(-3/2)#

#rArrf'(a)=f'(1)=1/2xx1/sqrt9=1/2xx1/3=1/6#

#rArr(f''(a))/(2!)=(f''(1))/2=1/2xx-1/4xx1/sqrt((9)^3)#

#color(white)(xxxxxxxx)=1/2xx-1/4xx1/27=-1/216#

#rArrf(x)=3+1/6(x-1)-1/216(x-1)^2#

#color(white)(rArrf(x))=3+1/6x-1/6-1/216x^2+1/108x-1/216#

#color(white)(rArrf(x))=-1/216x^2+19/108x+611/216larrcolor(red)"quadratic"#