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Using linear approximation to solve sqrt 49.4 as follows; let ƒ(x)= √x Equation of line is @ 49 can be written as y=mx+b where m is ? and b ? I found m which is 0.0714286*x+7-0.0714286*49 By doing this; ƒ'(x)=1/(2√x) ƒ'(49)=1/2√49 = a) L(x)= 7 + (1/14) x (x-49) L(49.4)=7 + (1/14) x (49.4-49) then I'm stuck at finding b What I did was take the sqrt of 49.4, which is 7.02851335632223 but the answer says error... Any help?
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Slope m = f'(49) = 1/14 => y = 1/14 ( x - 49) + 7 = x/14 + 7/2 Thus : y = 49.4/14 + 7/2 = 7.02857
y = 7.02857142857
Thanks Chlorophyll
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