Answer
($\frac{-c}{m}$+$\frac{1}{2}$, $\frac{c}{m}$+$\frac{1}{2}$) and 𝛿 =$\frac{c}{m}$
Work Step by Step
Solve the inequality |f(x) -L| < e
|mx+b-$\frac{m}{2}$ -b| 0;
Now we need to find the value of 𝛿>0 that places the open interval
($\frac{1}{2}$-𝛿, $\frac{1}{2}$+𝛿) centered at $\frac{1}{2}$ inside the interval($\frac{-c}{m}$+$\frac{1}{2}$, $\frac{c}{m}$+$\frac{1}{2}$). distance between $\frac{1}{2}$ and $\frac{-c}{m}$+$\frac{1}{2}$ is $\frac{c}{m}$ and distance between $\frac{1}{2}$ and$\frac{c}{m}$+$\frac{1}{2}$ is $\frac{c}{m}$. so we will take the smaller value for 𝛿, so 𝛿 =$\frac{c}{m}$
