1. ## Domain and range

Could you all please help me get started on finding the domain and range of these 2 functions

$g(x) = \sqrt{x^2-2x-5}$

and

$g(x) = (25+4x-x^2)^\frac{1}{2}$

Thank you

Update: i can find the domains, but the ranges are giving me trouble

2. Originally Posted by silencecloak
Could you all please help me get started on finding the domain and range of these 2 functions

$g(x) = \sqrt{x^2-2x-5}$

Mr F says: Let the domain be ${\color{red}(-\infty, ~ -a] \cup [b, ~ +\infty)}$. The function is decreasing over the first interval and increasing over the second ......

and

$g(x) = (25+4x-x^2)^\frac{1}{2}$

Mr F gives a hint: ${\color{red} y^2 = 25 + 4x - x^2 \Rightarrow x^2 - 4x + y^2 = 25 \Rightarrow (x - 2)^2 + y^2 = 29}$ ......

Thank you

Update: i can find the domains, but the ranges are giving me trouble
..

3. Originally Posted by mr fantastic
..
You've thoroughly confused me

4. Hello,

For the first one, the domain will be all x such that $x^2-2x-5 \ge 0$

$x^2-2x-5=x^2-2x+1-6=(x-1)^2-6$

$x^2-2x-5 \ge 0 \Longleftrightarrow (x-1)^2 \ge 6$

Can you solve it ?

Now the range...

$g(x) = \sqrt{\dots} \ge 0$

Try to continue

5. Quote:
Originally Posted by silencecloak
Could you all please help me get started on finding the domain and range of these 2 functions

Mr F says: Let the domain be . The function is decreasing over the first interval and increasing over the second ......

and

Mr F gives a hint: ......

Thank you

Update: i can find the domains, but the ranges are giving me trouble

Originally Posted by silencecloak
You've thoroughly confused me
There is a certain pre-requisite knowledge that a student working on questions like these is presumed to have.

You said you could get the domains. If that's the case, then you should know the value of a and b in . From how you got the domain it should be clear that the value of g starts at zero. If you stop and think about the behaviour of g (and keep in mind my earlier remarks about g decreasing and decreasing) I hope it's clear that the range must be [0, oo). Try drawing a graph of g to see this.

For the second, the graph of g(x) is a semi-circle. It's the upper half of the circle $(x - 2)^2 + y^2 = 29$. Draw the semi-circle, get the range by inspection.

6. Originally Posted by Moo
Hello,

For the first one, the domain will be all x such that $x^2-2x-5 \ge 0$

$x^2-2x-5=x^2-2x+1-6=(x-1)^2-6$

$x^2-2x-5 \ge 0 \Longleftrightarrow (x-1)^2 \ge 6$

Can you solve it ?

Now the range...

$g(x) = \sqrt{\dots} \ge 0$

Try to continue
am i suppose to put the equation under the square root sign?

and i was able to find the domains of both, by setting them to 0

i just dont know the method for the range

7. Originally Posted by mr fantastic
Quote:
Originally Posted by silencecloak
Could you all please help me get started on finding the domain and range of these 2 functions

Mr F says: Let the domain be . The function is decreasing over the first interval and increasing over the second ......

and

Mr F gives a hint: ......

Thank you

Update: i can find the domains, but the ranges are giving me trouble

There is a certain pre-requisite knowledge that a student working on questions like these is presumed to have.

You said you could get the domains. If that's the case, then you should know the value of a and b in . From how you got the domain it should be clear that the value of g starts at zero. If you stop and think about the behaviour of g (and keep in mind my earlier remarks about g decreasing and decreasing) I hope it's clear that the range must be [0, oo). Try drawing a graph of g to see this.

For the second, the graph of g(x) is a semi-circle. It's the upper half of the circle $(x - 2)^2 + y^2 = 29$. Draw the semi-circle, get the range by inspection.
I'm studying for a multiple choice timed test, i wont have time to graph and inspect.

8. Originally Posted by silencecloak
I'm studying for a multiple choice timed test, i wont have time to graph and inspect.
If you want to get the right answer I'd advise that you do graph and inspect .....