Answer
($\frac{-0.05}{m}$+1, $\frac{0.05}{m}$+1) and 𝛿 =$\frac{0.05}{m}$
Work Step by Step
Solve the inequality |f(x) -L| < e
|mx+b-m -b| <0.05
|m(x-1)| <0.05
-0.050;
Now we need to find the value of 𝛿>0 that places the open interval
(1-𝛿, 1+𝛿) centered at 1 inside the interval ($\frac{-0.05}{m}$+1, $\frac{0.05}{m}$+1). distance between 1 and $\frac{-0.05}{m}$+1 is $\frac{0.05}{m}$and distance between 1 and $\frac{0.05}{m}$+1 is$\frac{0.05}{m}$. so we will take the smaller value for 𝛿, so 𝛿 =$\frac{0.05}{m}$
