How do solve this problem (Finding deltas algebraically)?

[mighty canopy]

I am quite stuck and I am unsure. Please help.

[mint hound]

The main thing is just to unravel the mess of variables they've given. You're trying to solve for δ, so the first step is to write out the full system of inequalities. Then you can begin solving.

[mint hound]

.

[mighty canopy]

what does that mean?

[mint hound]

well you have a bunch of realtions, eg: ε = c < 0

[mighty canopy]

mx+b-((m/2)+b)
mx-m/2 < c>0

[mint hound]

if you list all the relations (ideally vertically), you can start to rearrange or replace some of them until you have δ bound between some values

[mighty canopy]

ok

I am still confused sorry

[mint hound]

write out what you have, then report back

[mighty canopy]

alright

[mighty canopy]

f(x) = mx+b, m > 0, L=(m/2) +b, c=1/2, ε=c>0
|f(x)-L| < ε
|mx+b-(m/2+b)|< (c>0)
|mx+b-m/2-b| < (c>0)
|mx-m/2| < (c>0)
-(c>0 + m/2)<|mx-m/2|<(c>0+m/2)
-(c>0+m/2)<|mx|<(c>0+m/2)

I am kinda stuck at this point

[mint hound]

well, for one, you know L = m/2 + b and f(x) = mx + b, so since |f(x)-L| < ε, |mx + b - m/2 - b| < ε, which simplifies to m|x - 1/2| < ε

since f(x) and L don't appear anywhere else, you can remove L=(m/2) +b and f(x)=mx+b, and replace |f(x)-L| < ε with m|x - 1/2| < ε

nvmnd, just saw that you aready did that lol

[mint hound]

hmm, this doesn't seem to have everything

[mint hound]

you should have this (I just listed all the relations in the question)

[mighty canopy]

I am unsure what you mean?