Question

f(x)=2*sqrt(x-1) and g(x)=sqrt(x-1)

compute the functions and find domains of the following:

1) f(x)+g(x)
2) f(x)-g(x)
3) f(x)*g(x)
4) f(x)/g(x)

Answer 1

Replace sqrt(x-1) with "y" and everything is trivial, except one that results in y squared and will need to be evaluated. Assuming these are real functions, where is the sqrt function defined?

Answer 2

\[f(x)+g(x)=2 \sqrt{x-1}+\sqrt{x-1} \]
f+g will be continuous when f and g are continuous
x-1 has to be positive or zero, right?

Answer 3

yes right myininaya.
f(x)+g(x)=3sqrt(x-1) (+inf,1)right?

Answer 4

\[x-1 \ge 0\]
\[ x \ge 1 \]

Which it around
(1,inf) since inf>1

Answer 5

And also include 1
so we say [1,inf)

Answer 6

no no wait wait. [1,inf) means 1 is included and inf is not. but why don't inf?

Answer 7

because inf is not actually a number

Answer 8

also \[\sqrt{1-1}=\sqrt{0}=0 \text{; 0 exists!}\]

Answer 9

so we include 1

Answer 10

oh I see. so thus inf can never be a domain of any kind of function. right??

Answer 11

Right when we say [1,inf)
this just means we want to include 1 and every single number greater than 1
We never include infinity because it is not a number

Answer 12

ok ok. fine..
2)sqrt(x-1) domain:[1,inf)
3)2x-2 what would be this domain?

Answer 13

OK. This makes sense.

Answer 14

\[f(x) \cdot g(x)=2 \sqrt{x-1} \cdot \sqrt{x-1}\]
Domain here is still the same
We want to \[\sqrt{x-1} \text{ to be a real number}\]
In order to have that that is a real number we must again have that \[x-1 \ge 0\]
Which means \[x \ge 1 \]
Which in interval notation that means [1,inf)

Answer 15

no no. look. why domain remain same?? this function is the multiple of two different function. so its domain different other than f(x)+g(x) and f(x)-g(x). explain.

Answer 16

We want both f and g to be real right?

Answer 17

ok. so we look individually though we are computing any function?

Answer 18

4)2 domain plz...

Answer 19

ok so 4
4)
this time we have the \[\sqrt{x-1}\]
on the bottom

Answer 20

yes.

Answer 21

\[\frac{f(x)}{g(x)}=\frac{2 \sqrt{x-1}}{\sqrt{x-1}}\]

We don't want the bottom to be zero

Answer 22

so we don't want to include 1

Answer 23

but we also don't want x-1<0
we want x-1>0
which means x>1
which means in interval notation we have (1,inf)

Answer 24

so it can never be 1 and any -ve value. right?

Answer 25

yes (1, inf)

Answer 26

:)

Answer 27

ok w8 teacher. thnx..

Answer 28

this question was from Howard Anton calculus. But book's answer is different.

Answer 29

awesome stuff :)

Answer 30

teacher. are you there?

Answer 31

you have a question?

Answer 32

yes.