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what is the error in linear approximation when f(x)=sqrt(3+x) a=22 and delta x= .5
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First of all, what's the linear approximation of your function when a=22?
5+1/10(x-22)??
The error is given by delta(x)*(dy/dx at x=a).
so it would be (.5)*5?
no thats not right how do i find derivative at x=a?
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First, find f'(x), then substitute x=a there. :D
i got 1/10 for f^1 of (x)
Then just multiply that by 0.5 and that's the error. :D
f(a+.5) - the linear approx of f(a+.5)
f(x)=sqrt(3+x) a=22 and h = .5 f(a) = sqrt(25) = 5 f'(a) = 1/2sqrt(3+a) = 1/10 y = 1/10(22+.5-22) + 5 y = 5.05 f(25.5) - 5.05 = error
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