What is the domain of this function? I already tried (-inf,-2) it was wrong.

Mathematics

1 answer:

valkas [14]3 years ago

6 0

Hello!

First off, (-inf, -2) was half of the domain, so you were right on track, but almost there!

Since this function has a vertical asymptote at x = -2, any x-values that are equal to -2 cause the function to be undefined. So, we show that as (-∞, -2) because this function is a continuous function from that interval.

Since this function is also continuous from the interval -2 to ∞, we show that as the second part of our domain; written as (-2, ∞).

Remember that parentheses and brackets have different meanings when using them to state the domain/range of a function. Parentheses are used to <u>not</u> include that value, while brackets are used <u>to</u> include it.

In that case, we need to combine this two intervals using the "union" symbol, which is "U".

Therefore, the domain of the function is (-∞, -2) U (-2, ∞).

You might be interested in

Which two points on the number line represent numbers that can be combined to make zero? A B C D -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1

maria [59]

Answer: B and D

Step-by-step explanation:

B=-2

D=2

The negative and positive signs cancel out, leaving you with zero.

Ex: you take two steps back, but then you take two steps forward so you're back where you started (0).

8 0

3 years ago

I don’t know how to do this

stich3 [128]

To check for continuity at the edges of each piece, you need to consider the limit as $x$ approaches the edges. For example,

$g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}$

has two pieces, $2x+5$ and $x^2-10$ , both of which are continuous by themselves on the provided intervals. In order for $g$ to be continuous everywhere, we need to have

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)$

By definition of $g$ , we have $g(-3)=2(-3)+5=-1$ , and the limits are

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1$

$\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1$

The limits match, so $g$ is continuous.

For the others: Each of the individual pieces of $f,h$ are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.