I have the given problem:
√(2x+3)
I found the point of discontinuity which is x cannot equal -1.5
However, I'm not sure how to find the point(s) of continuity.
Is there a formula to find the discontinuity and continuity?
$\endgroup$ 13 Answers
$\begingroup$To find the points of continuity, you simply need to find the points of discontinuity take their difference with respect to the reals.
For example, if you are dealing with a rational expression, a point of discontinuity would be anywhere where the function would not be defined, namely where the denominator is equal to zero. The function would then be continuous for all values such that the denominator is non-zero.
Looking at your problem, the number under the square root must be greater than or equal to zero (if we restrict ourselves to the reals).
Hence to be defined, we want $ 2 x + 3 \geq 0 $, solving this gives us $x\geq-\frac{3}{2}$.
What does this mean? Namely for any $x\geq-\frac{3}{2}$ the function is continuous, so it is continuous on $[-\frac{3}{2} , \infty)$. Hence we can see the function would be discontinuous whenever $x<-\frac{3}{2}$, as the expression under the radial would be negative. Your answer only found the transition point between then the function went from being undefined to defined.
Overall your points of discontinuity are all the points in the interval $(-\infty,-\frac{3}{2})$, and is continuous on the interval $[-\frac{3}{2} , \infty)$.
[Notice when $x=-\frac{3}{2}$, you have $\sqrt{2\left(\frac{-3}{2}\right)+3}=\sqrt{0}=0$, so the point $x=-\frac{3}{2}$ is NOT a point of discontinuity]
$\endgroup$ 1 $\begingroup$Recall that a function is called continuous at a point $x_0\in[a,b]$ if:$\forall\epsilon>0\quad\exists\delta>0\quad\forall x\in[a,b]\quad\delta>|x-x_0|>0\quad\rightarrow\quad\epsilon>|f(x)-f(x_0)|.$
You can verify that this condition, called the Cauchy criterion, holds for every $x\ge-\frac{3}{2}$, and in addition see that for $x<-\frac{3}{2}$, $f(x)$ isn't defined, so the condition doesn't hold.
Verification is simple: you must find $\delta$ as a function of $\epsilon$, so that every time you have some $\delta,$ you can deduce from it an $\epsilon$ that satisfies the inequality.
Another approach is to note that $2x+3$ is continuous on $\mathbb{R}$ and $\sqrt{x}$ is continuous on $\mathbb{R}^+$,so their composition is continuous on the domain of definition.
$\endgroup$ $\begingroup$HINT: Consider the sets where each function is discontinuous, and look at the composition of these functions.
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