Question

find the domain of f(x)=[sqrt(x-4)]/ln(3-2x)+1 please help! test tomorrow!!

Answer 1

The sqrt of negative numbers and logs of zero and less are excluded from the domain.

Answer 2

Peachy nailed it.

Answer 3

Assumption: The function meant is \[\frac{ \sqrt{x-4} }{ \ln(3-2x) } + 1,\]not with the +1 on the denominator.

Complete Solution: The argument of the square root must be non-negative. Therefore, \[ x - 4 \ge 0, \]which gives \[ x \ge 4. \]Similarly, the argument of the natural logarithm must be positive, so we have: \[ 3 - 2x > 0 \rightarrow 2x < 3 \rightarrow x < \frac{3}{2}. \]There are no values of x that are great than or equal to 4 or less than 3/2. Therefore, the domain is the null set: \[ \boxed {\cancel{0}}. \]

Note: No Calculus required!

Answer 4

SINCE THE DOMAIN OF THE DENOMINATOR(-infinity,3/2) HAS NO POINT WHERE IT INTERSECTS THE DOMAIN OF THE NUMERATOR[4,infinity) ,the function has no ultimate domain.