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

Answer 1

Evelyn Agard

#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#

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Answer 2

Samantha Adkison

To find the linearization at ( a = 0 ) of ( f(x) = \frac{1}{\sqrt{2+x}} ), we first find the derivative of ( f(x) ) at ( x = 0 ). The derivative ( f'(x) ) can be calculated using the chain rule, and then evaluated at ( x = 0 ). The linearization ( L(x) ) at ( a = 0 ) is given by the equation:

[ L(x) = f(a) + f'(a)(x - a) ]

After substituting ( a = 0 ) and ( f(x) = \frac{1}{\sqrt{2+x}} ), along with the derivative ( f'(x) ) evaluated at ( x = 0 ), we can simplify to obtain the linearization at ( a = 0 ) of the function ( f(x) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.