How to Verify Linear Approximations and Find Accurate Values for x?

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Homework Statement


Verify the given linear approximation at
a = 0.
Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.)

1/(1 + 3x)^3 ≈ 1 − 9x

Homework Equations



I have no idea how to find the x values

The Attempt at a Solution



(1/(1+3x)^3)-0.1<1-9x<(1/(1+3x)^3)+0.1 is how my textbook sets up the problem and then all of a sudden they seem to compute the x values that is unknown to me. is it simply just solving the inequality for x (subtract 1 and then divide by -9 having x=0) thanks and any help will be appreciated.
 

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benEE2018 said:
is it simply just solving the inequality for x (subtract 1 and then divide by -9 having x=0)
Yes, it is just a matter of solving each inequality for x, but not setting x = 0 anywhere.
E.g. for (1+3x)-3-0.1<1-9x, solve the equation (1+3x)-3-0.1=1-9x to find the end of the interval. Of course, solving a cubic is non-trivial, so you should use a suitable approximation.
Solving each cubic will give you one or three real values. (Two is theoretically possible, but most unlikely.) You then need to figure out which ones are relevant and which side of them satisfies the inequality.
 
haruspex said:
Yes, it is just a matter of solving each inequality for x, but not setting x = 0 anywhere.
E.g. for (1+3x)-3-0.1<1-9x, solve the equation (1+3x)-3-0.1=1-9x to find the end of the interval. Of course, solving a cubic is non-trivial, so you should use a suitable approximation.
Solving each cubic will give you one or three real values. (Two is theoretically possible, but most unlikely.) You then need to figure out which ones are relevant and which side of them satisfies the inequality.

Actually, you get quartics, because
[tex]\frac{1}{(1+3x)^3}\pm 0.1 = 1-9x \Longrightarrow 1 \pm 0.1(1+3x)^3 = (1+3x)^3(1-9x)[/tex]