(e^( cos^(-1) (log x^2) ))^1/2

The base of log is 4

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

The base of log is 4 - Sep 24th 2009, 09:41 PMProve It
- Sep 24th 2009, 09:44 PMutsav
- Sep 24th 2009, 09:59 PMProve It
$\displaystyle f(x) = \sqrt{e^{\arccos{[\log{(x^2)}]}}}$.

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

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

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

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

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