Finding delta for |f(x) - 3| < 1/100 when f(x) = x² + x + 1

[zeion]

Homework Statement

Suppose f(x) = x² + x + 1, a = 1, and L = 3. Find a value d > 0 such that 0 < |x - a| < d implies |f(x) - L| < 1/100

Homework Equations

The Attempt at a Solution

Given 0<|x-1|<d implies 0<|x² + x + 1 - 3|<1/100

0< x² + x + -2 <1/100
0<(x+2)(x-1)<1/100

Assume
0<|x-1|<1
1<x<2
3<x+2<4

Then
3|x-1| < (x+2)|x-1| < 4|x-1| < 1/100

We need
4|x-1| < 1/100
|x-1| < 1/251/25 < 1, therefore d should be 1/25.

 

[n!kofeyn]

Assume
0<|x-1|<1
1<x<2
3<x+2<4

In the second step it should be 0<x<2.

 

[zeion]

Ok.
But that doesn't really affect the rest does it?

 

[n!kofeyn]

The 3 should be a 2 in the next two statements following that. Also, d should be 1/400 because you divide 1/100 by 4. You accidently multiplied by 4. Other than that, it looks good.

 

[zeion]

Oh oops lol.
Ok thanks.