Mathematics
OpenStudy (anonymous):

Use differentials or linearization to approximate the value shown cubed radical(26.1)?

OpenStudy (anonymous):

you want an approx of $\sqrt[3]{26.1}$?

OpenStudy (anonymous):

yes like the way you do it without a calculator

OpenStudy (anonymous):

$\sqrt[3]{26.1}\approx\sqrt[3]{27}=3$

OpenStudy (anonymous):

With f dened to be f(x) = x^(1/3), the goal is to approximate f(26.1) using a linearization of f. Such a linearization requires the choice of base point a (where the tangent line is drawn). The strategy for picking a is to choose it close to 26.1 so that the approximation will be good, but also choose it to be a number at which f and f0 can be easily evaluated. We choose a = 26. The corresponding linearization is L(x) = 26^(1/3) + (x-8)*(1/3)*(26)^(-2/3) Therefore, L(26.1) = 26^(1/3) + (x-26)*(1/3)*(26)^(-2/3) = 2.96 that is almost the same as 26.1^(1/3)

OpenStudy (anonymous):

yea i got that and its wrong for some reason. round to four decimal places so i put 2.9653 and it still doesnt work

OpenStudy (anonymous):

I have the 2 values righ here in my calculator and its a very good aproximation: L(26.1) = 26^(1/3) + (x-26)*(1/3)*(26)^(-2/3) = 2.96629 and 26.1^(1/3) = 2.96628

OpenStudy (anonymous):

yea i have tried 2.9662 as well...

OpenStudy (anonymous):

but what is your problem? I can't figure out. Your question is answered lol L(26.1) = 26^(1/3) + (26.1-26)*(1/3)*(26)^(-2/3) = 2.96629

OpenStudy (anonymous):

i know but when i entered that number and four other numbers close to it i got it wrong...i only get five tries. but its fine.