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4 years ago

WILL GIVE BRAINLIEST

SAT

1 answer:

WARRIOR [948]4 years ago

6 0

Answer:

Parents who don't want their children dating set this restriction for different reasons, most of them are either an irrational fear of something happening to their child (which really just hold the child back because, whether you like it or not things will happen) or because of religious reasons (in that case.. you shouldn't date).

Most parents just want what is best for their child and even more so when they're "protecting" their only child. The problem with this is, through dating you learn what you like and don't like. You learn what you want and how to deal with certain things. Some parents tend to hold their children back from seeing the "dangerous side of the world" only for them to be thrown into it when they're adults and living on their own. In this case, maybe the mother wanted her son to reach a certain age before dating, or she wants him to date a person he can see versus online dating.

Hope this helps,

❤A.W.E.S.W.A.N.❤

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The functions $f$ and $g$ are defined as follows: \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}

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Recall that √x has a domain of x ≥ 0.

So, f(x) is defined as long as

(x + 1)/(x - 1) ≥ 0

• We have equality when x = -1

• Otherwise (x + 1)/(x - 1) is positive if both x + 1 and x - 1 are positive, or both are negative:

$\begin{cases}x+1>0 \implies x>-1 \\ x-1>0 \implies x>1\end{cases} \implies x > 1$

$\begin{cases}x+1$

Then the domain of f(x) is

x > 1 or x ≤ -1

On the other hand, g(x) is defined by two individual square root expressions with respective domains of

• x + 1 ≥ 0 ⇒ x ≥ -1

• x - 1 ≥ 0 ⇒ x ≥ 1

but note that g(1) is undefined, so we omit it from the second domain.

Then g(x) is defined so long as both x ≥ -1 *and* x > 1 are satisfied, which means its domain is

x > 1

f(x) and g(x) have different domains, so they are not the same function.

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