Question

please help!
Domain and range of:

a. f(x)= 1/x+3

b. g(x)= sqrt(x+6)

Answer 1

also h(x)= x^3+2x+5

Answer 2

is it
\[\frac{ 1 }{ x + 3 }~~~~~or ~~~~~~\frac{ 1 }{ x } + 3\]

Answer 3

if its \[\frac{ 1 }{ x + 3 }\]
then the denominator can't be zero.
i.e.\[x + 3 \neq 0\]
so \[x \neq -3\]

Answer 4

ok thanks! what about range? and the other ones

Answer 5

@rishavraj

Answer 6

did u get the domain ????

Answer 7

for which one?

Answer 8

i got the first one! @rishavraj

Answer 9

and i think i got sqrt(x+3) but i'm not sure

Answer 10

the functn in sqrt can't be negative....
so
\[x + 3 \ge 0\]

Answer 11

i got x≥-3

Answer 12

oops the question was actually sqrt (x+6) so x≥-6 ?

Answer 13

or is it x>6? because couldn't it be sqrt(6-6)

Answer 14

its \[x \ge -6\]

Answer 15

oh oops i meant -6, but still why isn't it x>-6

Answer 16

you looking for the range of
\[f(x)=\frac{1}{x+3}\]?

Answer 17

yes please

Answer 18

i got all real numbers but not sure

Answer 19

a fraction is only zero if the numerator is zero

Answer 20

therefore
\[\frac{1}{x+3}\] cannot be zero because the numerator is 1

Answer 21

range is all real numbers except zero

Answer 22

oh! oops. thanks! what about sqrt(x+6) and h(x)=x^3+2x+5

Answer 23

anyone?

Answer 24

its clear the its domain and range is ALL REAL Numbers... i.e R

Answer 25

which one are we talking about :/

Answer 26

\[\sqrt{x+6}\] the square root of anything is never less that zero
range
\[y\geq 0\]

Answer 27

about h(x) = x^3 + 2x + 5

Answer 28

cube goes from \(-\infty\) to \(+\infty\)

Answer 29

oh, ok thanks! and im still confused about the sqrt one?

Answer 30

\[y\geq 0\]

Answer 31

why isn't it y>0?

Answer 32

bcoz it can be even zero thts why.....

Answer 33

oh. why is the domain x≥-6

Answer 34

i thought it could be sqrt0 or no

Answer 35

see the function in sqrt can never be negative...... so
\[x + 6 \geq 0\]
\[x \geq -6\]

Answer 36

oh okay thank you so much!