+23245 In #28:
ln(x² - 1): ln(x) is defined only for positive values of x, so:
x² - 1 > 0 ---> x² > 1 ---> x > 1 or x < -1
In the denominator: x² - 2x must be positive:
x² - 2x > 0 ---> x(x - 2) > 0
either x > 0 and x - 2 > 0 or x < 0 and x - 2 < 0
---> x > 2 or x < 0
Now put the first two answers with the second two answers and take the most restrictive cases: x > 1 and x > 2 means x >2 X < -1 and x < 0 means x < -1
Therefore, it will be continuous in the region where x < -1 and again in the region where x > 2 (but not between these two regions).
+23245 I look at this problem this way:
y = invSin(x+2) ---> sin(y) = x + 2 ---> x = sin(y) - 2.
The sin is only defined for values at least as large as -1 and no larger than +1; so sin(y) is pinned in this way: -1 <= sin(y) <= 1; this means that x >= -3 and x <= -1.
+23245 In #28:
ln(x² - 1): ln(x) is defined only for positive values of x, so:
x² - 1 > 0 ---> x² > 1 ---> x > 1 or x < -1
In the denominator: x² - 2x must be positive:
x² - 2x > 0 ---> x(x - 2) > 0
either x > 0 and x - 2 > 0 or x < 0 and x - 2 < 0
---> x > 2 or x < 0
Now put the first two answers with the second two answers and take the most restrictive cases: x > 1 and x > 2 means x >2 X < -1 and x < 0 means x < -1
Therefore, it will be continuous in the region where x < -1 and again in the region where x > 2 (but not between these two regions).
+118609 No 26
f(x)=ln(sinx)
sinx must between -1 and 1
okay, ln(whatever) The 'whatever' must be greater than 0. You can only find the log of a positive number.
so ln(sinx) must be between 0 and 1 CORRECTION: So sinx must be between 0 and 1
x must be in the 1st or second quadrant.
How about
$$2n\pi
I invite othe mathematicians to examine what I have said. 
+1832 Melody, I don't understand the last part
x must be in the 1st or second quadrant.
+118609 Hi 315 
You have picked me up on an error - that is good
No 26
"f(x)=ln(sinx)
sinx must between -1 and 1
okay, ln must be greater than 0
so ln(sinx) must be between 0 and 1 WRONG sinx must be between 0 and 1.
------------------------------
I think I have made a 'wording' error here - It actually means that sinx must be between 0 and 1 because you cannot find the log of a negative number.
sin x is only positive in the first and the second quadrant. etc
Does that make sense now?
+118609 No,
sin 0 = 0
ln(0) is undefined
same goes with x=pi
So
ONE of the intervals is (0,pi)
But to get all of them i will redirect you to my first answer
