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