+2498 For how many positive integers, x, does the function:
\(f(x)=\frac{\sqrt{x-3}}{x^2-8x-20}\)
have no real values?
(A) 2
(B) 3
(C) 4
(D) Inf.
+129852 Actually...I believe the answer should be 3
We are only concerned with positive integer values for x
Clearly.....the posiitve integers 1 and 2 make the radical negative...and thus "not real'
And...x^2 - 8x - 20 = 0 factors as (x -10) (x + 2) = 0.....so the integers -2 and 10 make the denominator 0......but 10 is the only positve x that makes the denominator 0
So.......the positive integer values for x that make this function undefined are 1, 2 and 10
+23252 Functions will not have real answers if the denominator is zero.
Factoring the denominator: x2 - 8x - 20 = (x - 10)(x - 2) ---> therefore the two numbers, 10 and 2, must be excluded.
Also, functions will not have real answers if there is a negative number under the square root sign.
All numbers smaller than 3 will give a square root of a negative number, therefore, they must all be excluded. This is an infinite number of numbers.
The answer, then, is (d) -- you must exclude an infinite number of numbers.
+129852 Actually...I believe the answer should be 3
We are only concerned with positive integer values for x
Clearly.....the posiitve integers 1 and 2 make the radical negative...and thus "not real'
And...x^2 - 8x - 20 = 0 factors as (x -10) (x + 2) = 0.....so the integers -2 and 10 make the denominator 0......but 10 is the only positve x that makes the denominator 0
So.......the positive integer values for x that make this function undefined are 1, 2 and 10
