I need to find thelocal approximation of f(x)=x^(1/3) at x=26.6, knowing f(27)=3.(adsbygoogle = window.adsbygoogle || []).push({});

Here's what I did, dont know if I did it right:

f '(x)=(1/3)x^(-2/3)=[x^(-2/3)]/3

slope of the tanget = slope of the secant

[x^(-2/3)]/3=(y-y1)/(x-x1)

[26.6^(-2/3)]/3=(y-3)/(x-27)

now I sub in X and solve for y:

[26.6^(-2/3)]/3=(y-3)/(26.6-27)

y=2.985

I dont think that's the answer at all, could someone check it?

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# Linear Approximation

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