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anonymous
 4 years ago
21.) For what values of x may (x+1)^(1/3) be replaced by x^(1/3), if the allowable error is 0.01?
ans... x>192
22.) For what values of x may (x+1)^(1/4) be replaced by x^(1/4), if the allowable error is 0.01?
ans... x>73
anonymous
 4 years ago
21.) For what values of x may (x+1)^(1/3) be replaced by x^(1/3), if the allowable error is 0.01? ans... x>192 22.) For what values of x may (x+1)^(1/4) be replaced by x^(1/4), if the allowable error is 0.01? ans... x>73

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can anyone please help with the solution??? Thanks.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.121. Solve \[(x+1)^{1/3}x^{1/3}=0.01\] Depending what maths you're doing, you can solve the equation by trial and error or by Newton's method. You'll find that for x\( \le \)192, the lefthandside is greater than 0.01. So the answer is x>192. For #22, the process is the same.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks, but what's the solution using approximating formulas (Differentials)?

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1Thanks for providing the context. Let f(x)=x^(1/3) f'(x)=(1/3)x^(2/3) Using linearization, \( \Delta f = f'(x) \Delta x\) Put \( \Delta x=1\, \ and\ \Delta f = 0.01, \) we get \(f'(x)=\frac{1}{3}x^{\frac{2}{3}}=0.01/1=0.01 \) solving, \(x=(3*0.01)^{\frac{3}{2}}=192.45009\ (approx.)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot! We haven't discussed linearization yet though (I don't even know what it is yet) but I'll try to understand. Thanks!
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