1. ## Functions

The intersection of the domains of functions: $f(x) = arcsin(log\frac{x}{10})$ and $g(x) = \sqrt{2+x-x^2}$

a) $[1,2]$
b) $(0,2]$
c) $[1, \infty)$
d) $\phi$
e) $\mathbb{R}_+^*$

2. Originally Posted by Apprentice123
The intersection of the domains of functions: $f(x) = arcsin(log\frac{x}{10})$ and $g(x) = \sqrt{2+x-x^2}$

a) $[1,2]$
b) $(0,2]$
c) $[1, \infty)$
d) $\phi$
e) $\mathbb{R}_+^*$
1. $g(x) = \sqrt{2+x-x^2}$

Solve for x:

$2+x-x^2 \geq 0$

I've got $x \in [-1, 2]$

2. The domain of the arcsin-function is [-1, 1].

I assume that $\log$ means $\log_{10}$

$\log\left(\dfrac x{10}\right) = -1~\implies~x = 1$

$\log\left(\dfrac x{10}\right) = 1~\implies~x = 100$

That means the domain of the given arcsin-function is [1, 100]

3. Now calculate the intersection. (I've got answer a))

3. ## hi

Easiest is looking at the second function maybe.
For the expression under the square root sign to be positivt, we solve
$-x^2 + x +2 = (x-2)(x+1)$
So, this function makes boundaries $[-1,2]$

Looking at $arcsin(log(\frac{x}{10})) = arcsin(log(x) - log(10))$

That is, $arcsin(log(x) - 1)$

The logharitm is defined for $x>0$, so x>0
and arcsin(x) for $|x| \leq 1$ .

So we get $|log(x) - 1| \leq 1$ , giving us
when we combine the interval $[1,2]$

4. thank you