# Thread: Fidn the domain of..

1. ## Find the domain of..

Sqrt[ (1-x^2) / (ln(4-2x) ]

Okay.. So, since it is a square root, I know everything under the radical must be >= 0, so

Sqrt[ (1-x^2) / (ln(4-2x) ] >= 0

But, since it is a rational fraction, the demoninator can't be zero.

ln(4-2x) != 0

I also know since it is a LN, the input must be > 0.. so..

4-2x >= 0

Am I on the right start here, in discovering the domain?

4 - 2x >= 0
x < 2

ln(4-2x) != 0
4-2x != e^0
-2x != -4 + 1
-2x != -3
x != 3/2

this just leaves hte last one, i think.. can someone help me?

Thank you

2. Originally Posted by zodiacbrave
Sqrt[ (1-x^2) / (ln(4-2x) ]

Okay.. So, since it is a square root, I know everything under the radical must be >= 0, so

Sqrt[ (1-x^2) / (ln(4-2x) ] >= 0

But, since it is a rational fraction, the demoninator can't be zero.

ln(4-2x) != 0

I also know since it is a LN, the input must be > 0.. so..

4-2x >= 0

Am I on the right start here, in discovering the domain?

4 - 2x >= 0
x < 2

ln(4-2x) != 0
4-2x != e^0
-2x != -4 + 1
-2x != -3
x != 3/2

this just leaves hte last one, i think.. can someone help me?

Thank you
$\displaystyle f(x) = \sqrt{\frac{1 - x^2}{\ln{(4 - 2x)}}}$.

First, look for any restrictions that you know of.

You know that you can not have the logarithm of a nonpositive number, so to start with

$\displaystyle 4 - 2x > 0$

$\displaystyle 4 > 2x$

$\displaystyle x < 2$.

Also, since you have a square root, everything under the square root must be nonnegative.

So $\displaystyle \frac{1 - x^2}{\ln{(4 - 2x)}} \geq 0$.

Note that since the denominator is ALWAYS positive, the numerator can not be negative.

So $\displaystyle 1 - x^2 \geq 0$

$\displaystyle 1 \geq x^2$

$\displaystyle x^2 \leq 1$

$\displaystyle |x| \leq 1$

$\displaystyle -1 \leq x \leq 1$.

So we have in our domain

$\displaystyle x < 2$ AND $\displaystyle -1 \leq x \leq 1$.

For both inequalities to hold, $\displaystyle -1 \leq x \leq 1$.

Thus the domain of f is $\displaystyle -1 \leq x \leq 1$.

3. $\displaystyle \frac{3}{2} < x < 2$ is also in the domain.

$\displaystyle \ln(4-2x) < 0$ in that interval, as is $\displaystyle (1-x^2)$ , making $\displaystyle \frac{1-x^2}{\ln(4-2x)} > 0$

4. Originally Posted by skeeter
$\displaystyle \frac{3}{2} < x < 2$ is also in the domain.

$\displaystyle \ln(4-2x) < 0$ in that interval, as is $\displaystyle (1-x^2)$ , making $\displaystyle \frac{1-x^2}{\ln(4-2x)} > 0$
Edit - you are right...