University Calculus: Early Transcendentals (3rd Edition)

1) The open interval is $(1.703,1.761)$ 2) $\delta=0.028$
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-\sqrt3|\lt\delta\Rightarrow|(x^2)-3|\lt0.1$$ 1) Find the interval around $\sqrt3$ on which $|(x^2)-3|\lt0.1$ holds. Solve the inequality: $$|(x^2)-3|\lt0.1$$ $$-0.1\lt x^2-3\lt0.1$$ $$2.9\lt x^2\lt3.1$$ Now we need to take the square root. However, there are 2 cases, whether $x\ge0$ or $x\lt0$. - For $x\ge0$: $1.703\lt x\lt1.761$ - For $x\lt0$: $-1.761\lt x\lt-1.703$ Yet, here we are looking for the interval around $\sqrt3$, which means $x\gt0$. Therefore, we choose only the open interval around $\sqrt3$, which is $(1.703,1.761)$. 2) Give a value for $\delta$ - From $1.703$ to $\sqrt3$: $\sqrt3-1.703\approx0.0291$ - From $1.761$ to $\sqrt3$: $1.761-\sqrt3\approx0.0286$ The nearer endpoint to $\sqrt3$ is $1.761$, and the distance between them is about $0.0286$. So to make sure, if we take $\delta=0.028$ or any smaller positive number, then $0\lt|x-\sqrt3|\lt0.028$, meaning all $x$ would be placed in the interval $(1.703,1.761)$ so that $|(x^2)-3|\lt0.1$. In other words, $$0\lt |x-\sqrt3|\lt0.028\Rightarrow|(x^2)-3|\lt0.1$$