Asked by Anonymous
Use a linear approximation to estimate the number (27.07)^(2/3)
Answers
Answered by
Reiny
using:
f(x)≈f(xo)+f′(xo)(x−xo) <-- should be in your text
27.07 = 27 + .07
let f(x) = x^(2/3)
f'(x) = (2/3)x^(-1/3)
let xo = 27, x = 27.07
f(27.07) ≈ f(27) + f'(27)(27.07 - 27)
= 27^(2/3) + (2/3)(27)^(-1/3) (.07)
= 9 + (2/3)((1/3)(.07)
= 9 + .01555..
= 9.01555..
by calculator, 27.07^(2/3) = 9.0155488..
not bad !!
f(x)≈f(xo)+f′(xo)(x−xo) <-- should be in your text
27.07 = 27 + .07
let f(x) = x^(2/3)
f'(x) = (2/3)x^(-1/3)
let xo = 27, x = 27.07
f(27.07) ≈ f(27) + f'(27)(27.07 - 27)
= 27^(2/3) + (2/3)(27)^(-1/3) (.07)
= 9 + (2/3)((1/3)(.07)
= 9 + .01555..
= 9.01555..
by calculator, 27.07^(2/3) = 9.0155488..
not bad !!
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