Use algebrad the range of
i remember our teacher told us to solve for A, any real number so I did
A cannot equal 3
I don't know what the next step is
1. Since for all the LHS of the equation must be greater zero too.
2. A quotient is greater (or equal) zero if the signs of denominator and numerator are equal:
That's the range of the function.
Btw: Please write numerator and denominator of a quotient in brackets. Otherwise you mean:
Think of it this way. You found your critical points to be 3 and -10/9 by setting the numerator and denominator to 0
Now test a value from - infinity to -10/9, next a value from -10/9 to 3, and then from 3 to positive infinity.
You will notice that end up with a positive value in the intervals of (- infinity to -10/9) and (3 to positive infinity).
You will end up with a negative value in the interval of (-10/9 to 3)
Now since you can only have real solutions you can't use the negative interval.
The reason why you can't use the negative interval is when you take a square root of a negative number you get an imagnery number.
Does this make sense?
So basicly, 9A+10/A-3 cannot be a negative number because you can't square root a negative number and still have a real number, because of and -10/9 is the number that makes the fraction 0 and any number less than that would make the fraction negative correct?