please help! Domain and range of: a. f(x)= 1/x+3 b. g(x)= sqrt(x+6)

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please help! Domain and range of: a. f(x)= 1/x+3 b. g(x)= sqrt(x+6)

Mathematics
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also h(x)= x^3+2x+5
is it \[\frac{ 1 }{ x + 3 }~~~~~or ~~~~~~\frac{ 1 }{ x } + 3\]
if its \[\frac{ 1 }{ x + 3 }\] then the denominator can't be zero. i.e.\[x + 3 \neq 0\] so \[x \neq -3\]

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Other answers:

ok thanks! what about range? and the other ones
did u get the domain ????
for which one?
i got the first one! @rishavraj
and i think i got sqrt(x+3) but i'm not sure
the functn in sqrt can't be negative.... so \[x + 3 \ge 0\]
i got x≥-3
oops the question was actually sqrt (x+6) so x≥-6 ?
or is it x>6? because couldn't it be sqrt(6-6)
its \[x \ge -6\]
oh oops i meant -6, but still why isn't it x>-6
you looking for the range of \[f(x)=\frac{1}{x+3}\]?
yes please
i got all real numbers but not sure
a fraction is only zero if the numerator is zero
therefore \[\frac{1}{x+3}\] cannot be zero because the numerator is 1
range is all real numbers except zero
oh! oops. thanks! what about sqrt(x+6) and h(x)=x^3+2x+5
anyone?
its clear the its domain and range is ALL REAL Numbers... i.e R
which one are we talking about :/
\[\sqrt{x+6}\] the square root of anything is never less that zero range \[y\geq 0\]
about h(x) = x^3 + 2x + 5
cube goes from \(-\infty\) to \(+\infty\)
oh, ok thanks! and im still confused about the sqrt one?
\[y\geq 0\]
why isn't it y>0?
bcoz it can be even zero thts why.....
oh. why is the domain x≥-6
i thought it could be sqrt0 or no
see the function in sqrt can never be negative...... so \[x + 6 \geq 0\] \[x \geq -6\]
oh okay thank you so much!

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