How do you find the linearization at a=0 of #f(x)=1/1/(sqrt(2+x))#?

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Answer 1 · 2018-01-01T15:50:24

#p_1: y=sqrt2/2-(xsqrt2)/8#

#p_2: y=sqrt2+(xsqrt2)/4#

f(x)=1/1/(sqrt(2+x))

I don't know which one you meant. Here you go:

#y=f(a)+f'(a)(x-a)#

#f_1(x)=1/(sqrt(2+x))=(2+x)^(-1/2)#

#f(0)=1/(sqrt(2+0))=sqrt2/2#

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

#f'(0)=-1/(4sqrt2)=-sqrt2/8#

#p_1: y=sqrt2/2-(xsqrt2)/8#

#f_2(x)=sqrt(2+x)#

#f(0)=sqrt(2)#

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

#f'(0)=1/(2sqrt(2))=sqrt2/4#

#p_2: y=sqrt2+(xsqrt2)/4#