# Homework Help: For what values x is the linear approxiamation accurate to within .5

1. Jan 28, 2012

### bobsmith76

They're doing some algebra operation that I have forgotten how to do, maybe. I need to know what this operation is called so that I can go back in the book and find out how to do it. I don't see why they are putting √(x+3) on both sides of the equation.

2. Jan 28, 2012

### DivisionByZro

They're inequalities (and an absolute value). Basically when they write this:

$$\sqrt{x+3}-0.5 < \frac{7}{4}+\frac{x}{4} < \sqrt{x+3}+0.5\$$

They mean that $$\frac{7+x}{4}$$ is "within" the range ±0.5 of
$$\sqrt{x+3}$$
when "x" is some range or set of values.

In other words, think of it as an error in measurement of something. Ex: "This tube is 1mm wide, plus or minus 0.01mm". It's very similar, you could write the diameter of the tube almost in the same way.

So in reality they're asking you to solve these two inequalities:

$$\sqrt{x+3}-0.5 < \frac{7}{4}+\frac{x}{4}$$

and

$$\frac{7}{4}+\frac{x}{4} < \sqrt{x+3}+0.5\$$

for x. However, you need to find x that makes both true simultaneously. They seem to be doing it graphically. Does this clear things up a little?

3. Jan 28, 2012

yes, thanks.