Question

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

Answer 1

Can anyone please help with the solution??? Thanks.

Answer 2

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.

Answer 3

Thanks, but what's the solution using approximating formulas (Differentials)?

Answer 4

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

Answer 5

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!