What is the domain of f(g(x)) given f(x)= radical 25-x^2 and g(x)= ln(x+3)?

[Psichlohomeo]

1. Find the domain of f(g(x)) where f(x)= radical 25-x^2 and g(x)= ln(x+3)



2. For the domain of f I got x < or equal to +/- 5 and for the domain of ln(x+3) I got x+3 > 0



3. (0, 145)

Thanks for your time. :P

 

[willem2]

1. Find the domain of f(g(x)) where f(x)= radical 25-x^2 and g(x)= ln(x+3)



2. For the domain of f I got x < or equal to +/- 5 and for the domain of ln(x+3) I got x+3 > 0

Ol. I suppose you mean -5 <= x <= 5 for the domain of f.

3. (0, 145)

No.. The first number is wrong. the second number is nearly right, but since this is math and not physics: If you get e^5 in an anwer, just leave it.

 

[Psichlohomeo]

Ah. So, [e^-5, 0)(0, e^5]?

 

[Psichlohomeo]

Also, how can I find the domain of f o g when f(x) = log(x-1) and g(x)=x/16-x^2?

For the domain of f(x), I put x>2 (??) and g(x) cannot equal +/- 4. I got this strange answer: (-1+rad65/-2 , -4) (-4, -1-rad65/-2)(-1-rad65/-2 , 4)

Thank again. :)

 

[Bohrok]

log(x) is defined for x>0. The domain of log(x-1) would be where x-1>0
You got the domain of g(x) right.

$$f\circ g(x) = f(g(x)) = \log\left(\frac{x}{16-x^2}\right)$$

Putting the domains together, what will be the domain of the composition? Where will the x values be that make it defined?

 

[willem2]

Ah. So, [e^-5, 0)(0, e^5]?

there's no problem at 0. and you forgot about the x+3.

 

[Psichlohomeo]

Alright, for my first problem I've reworked it to (-3, infinity)