Find Interval for |f(x)-L|<E: -sqrt 4.5 < x < -sqrt 3.5

[kuahji]

Find an open interval about a on which the inequality | f(x) - L | < E holds. Then give a value for D > 0 such that for all x satisfying 0 < | x - a | < D the inequality | f(x) - L | < E holds.

f(x) = x^2, L = 4, a = -2, E = .5

To solve I setup the inequality

Step 1: -.5 < x^2 - 4 < .5

Step 2: 3.5 < x^2 < 4.5

This is where I get stuck, need help with the algebra.

I got sqrt 3.5 < x < sqrt 4.5 & thought my interval for E would equal (sqrt 3.5, sqrt 4.5), but the book gives (-sqrt 4.5, -sqrt 3.5). I know if you take the sqrt, you have to take the positive & the negative of that number, but how do you know which to choose in this case & by doing this, does the inequality signs always reverse (like < to >)? Going to skip the rest of the problem, because I didn't have a problem solving the rest.

 

[Dick]

Your answer is correct for a=+2, but the question says a=-2. Draw a graph of f(x)=x^2 and look around the points x=2 and x=-2. When you are taking a sqrt and you have choice of +/- you have to think about the problem.

 

[HallsofIvy]

Your problem is not to find bounds on x² or x but on x- a which, here, is x+2.
From -.5< x²-4< .5, factor to get -.5< (x-2)(x+2)< .5 and then think about dividing by x-2. With x "close" to -2, say between -1 and -3, what is the largest x-2 can be? What is the smallest?