Use the alternating series estimation theorem to estimate the range of values of x for which the given approximation is accurate to within the stated error. check your answer graphically. (round your answers to three decimal places.) sin(x) ≈ x − x3 6 |error| < 0.000001

The expansion series of sin(x) about 0 is
  ...............(1)

Using the given approximation S(x)=x-x^3/3!           ................(2)
Therefore the error term for the approximation is 
error=(2)-(1)



The alternative series theorem says we need only to ensure that the absolute value of the first term dropped (x^5/5!) is less than the error limit, in order to ensure that the sum will be below the error limit
This means that
|x^5/5!| <  0.000001

Solving for x:
x^5=5!*0.000001=0.000120
x=(0.000120)^(1/5)=0.164375

Thus the approximation of sin(x) ≈ x-x^3/3! has an absolute error below 0.000001 for the interval [-0.164375,+0.164375].

Check:
sin(0.164375)-(0.164375)^3/3!=9.993513*10^(-7) < 0.000001

Dang i wish i could , good luck (i’m bad at math)