# The function is defined if...

• Sep 24th 2009, 09:39 PM
utsav
The function is defined if...
(e^( cos^(-1) (log x^2) ))^1/2

The base of log is 4
• Sep 24th 2009, 09:41 PM
Prove It
Quote:

Originally Posted by utsav
(e^( cos^(-1) (log x^2) ))^1/2

The base of log is 4

What are you trying to do?
• Sep 24th 2009, 09:44 PM
utsav
Quote:

Originally Posted by Prove It
What are you trying to do?

The question is: The function is defined if..., we have to find the domain of the given function.
• Sep 24th 2009, 09:59 PM
Prove It
Quote:

Originally Posted by utsav
(e^( cos^(-1) (log x^2) ))^1/2

The base of log is 4

$f(x) = \sqrt{e^{\arccos{[\log{(x^2)}]}}}$.

Note that the logarithm is only defined for $x > 0$.

Since $x^2 \geq 0$, we can already see that we can not let $x = 0$.

Also note that $\arccos{x}$ is only defined for $-1 \leq x \leq 1$.

The exponential function is always $>0$, so there are not any further restrictions for the square root.

Thus the domain is $x \in [-1, 0)\cup(0, 1]$.