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- anonymous

Give and example ( by writing an equation) of each of the following:
a function whose domain is [0, infinity)
a function whose domain is ( -infinity, infinity)
a function whose domain is (-infinity, 0) u ( 0, infinity)

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Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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- anonymous

- schrodinger

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- anonymous

help

- anonymous

- anonymous

help

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- anonymous

@IrishBoy123 some one help i give medals!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

- anonymous

ill help

- anonymous

thanks

- anonymous

one sec

- anonymous

k

- anonymous

f(x) = x + 1 for number 2

- anonymous

f(x) = x + 1 for number 2

- anonymous

1-x if x is greater than or equal to zero. third 1

- anonymous

and finally the first 1 is f(x)= x +4 if x is less than zero

- anonymous

did this help @Julianne6th

- anonymous

yes

- anonymous

is this right

- anonymous

You can't take the square root of negative numbers, so the domain of \[f(x) = \sqrt x\] is: \[[0,\infty)\]

- anonymous

What is the domain of this function: \[f(x) = 1/x\]
(Think: what number(s) are you not allowed to plug in for x?)

- anonymous

this is @cookiimonster627 question i posted it for him or her

- anonymous

just FYI: the domains of both f(x) = 1-x and f(x) = x + 4 are all real numbers: that is, the interval (-infinity,infinity).
Why? What number(s) can be plugged into 1 - x ? Any numbers!
That is, you can subtract *any* number from 1. (and so on)

- anonymous

@OneMathCat your so wrong that shows your not a math cat

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