## 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

1. anonymous

2. mathmate

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.

3. anonymous

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

4. mathmate

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

5. anonymous

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!