Question

Determine the restrictions for f(x) = log(25-x^2)

Answer 1

x<5

Answer 2

How did you get that sir?

Answer 3

or |x| < 5

Answer 4

hmm you have to make sure that
\[x^2-25>0\] right

Answer 5

how would i show my work for that?

Answer 6

sorry i meant
\[25-x^2>0\] so
\[(5+x)(5-x)>0\] meaning
\[-5<x<5\]

Answer 7

im confused...

Answer 8

The justification is because log(x) does not take zero or negative arguments.

Answer 9

\[y=25-x^2\] is a parabola that opens down. it is therefore positive between the zeros and negative outside them. you cannot take the log of a non-positive number.
see lousy picture

Answer 10

haha its ok, I understand but i still dont know how to show my work

Answer 11

you don't think my picture is good enough?

Answer 12

its great :)

Answer 13

what are the steps to how you would come to the conclusion that x<5?

Answer 14

Can anyone please help?

Answer 15

If Satellite can't help you, not sure if too many would succeed.

I'll give it another try:
It all summarizes in two facts you have to establish.
1. The domain of log(x) is (0,+inf), so it does not admit zero or negative values of x.
2. To achieve that, as explained by Satellite, you need
25-x^2 >0, or
x^2<25, or
-5<x<+5.
That's about it.

Answer 16

be careful. it is not
\[x<5\] as you stated, but rather what mathmate said
\[-5<x<5\] this is because on that interval
\[25-x^2>0\] and so
\[\log(25-x^2)\] is defined