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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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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Can anyone please help with the solution??? Thanks.
21. 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 left-hand-side is greater than 0.01. So the answer is x>192. For #22, the process is the same.
Thanks, but what's the solution using approximating formulas (Differentials)?

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Thanks 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, \(x=(3*0.01)^{-\frac{3}{2}}=192.45009\ (approx.)\)
Thanks 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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