+0

Again !

+3
921
4
+2533

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.

Mar 18, 2016

#3
+116126
+6

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

Mar 18, 2016

#1
+21972
+5

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.

Mar 18, 2016
#2
+2533
0

Thank you Geno3141 ! :)

Solveit  Mar 18, 2016
#3
+116126
+6

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

CPhill Mar 18, 2016
#4
+2533
0

Yes you are right thanks CPhill ! :)

I understood the general idea

Solveit  Mar 18, 2016