University Calculus: Early Transcendentals (3rd Edition)

1) The open interval is $(1/2-\frac{c}{m},1/2+\frac{c}{m})$ 2) $\delta=\frac{c}{m}$
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-\frac{1}{2}|\lt\delta\Rightarrow|(mx+b)-(\frac{m}{2}+b)|\lt c\hspace{1cm}m\gt0, c\gt0$$ 1) Find the interval around $\frac{1}{2}$ on which $|(mx+b)-(\frac{m}{2}+b)|\lt c$ holds. Solve the inequality: $$|(mx+b)-(\frac{m}{2}+b)|\lt c$$ $$-c\lt (mx+b)-(\frac{m}{2}+b)\lt c\hspace{1cm}(c\gt0)$$ $$-c\lt mx+b-\frac{m}{2}-b\lt c$$ $$-c\lt mx-\frac{m}{2}\lt c$$ $$-\frac{c}{m}\lt x-\frac{1}{2}\lt\frac{c}{m}\hspace{1cm}(m\gt0)$$ $$\frac{1}{2}-\frac{c}{m}\lt x\lt\frac{1}{2}+\frac{c}{m}$$ The open interval around $1/2$ is $(1/2-\frac{c}{m},1/2+\frac{c}{m})$. 2) Give a value for $\delta$ Both endpoints are equally distant from $1/2$, the distance of which is $\frac{c}{m}$ So if we take $\delta=\frac{c}{m}$ (since $m\gt0$ and $c\gt0$, $\delta=\frac{c}{m}\gt0$) or any smaller positive number, then $0\lt|x-1/2|\lt\frac{c}{m}$, meaning all $x$ would be placed in the interval $(1/2-\frac{c}{m},1/2+\frac{c}{m})$ so that $|(mx+b)-(\frac{m}{2}+b)|\lt c$. In other words, $$0\lt |x-1/2|\lt\frac{c}{m}\Rightarrow|(mx+b)-(\frac{m}{2}+b)|\lt c$$