Find the domain of the function and express it in interval notation.

Question

ƒ(x)= 1 / √4x² - 7x + 3

 

(^^^ That's a square root symbol ^^^)

 

Have tried many times with no correct answers so far

 

Please help ASAP!

 

Thanks :)

Answer 1

I see that this has been answered twice.  I don't know if you know how

the two previous tutors factored the expression ... Also, there is an easier

and logical way to work the inequality..

 

ƒ(x)= 1 / [√(4x² - 7x + 3)]

 

Since we need positive numbers under the square root to have

real answers and we cannot divide by zero, the domain depends on

 

4x²-7x+3>0

 

We can factor by grouping:  Multiply the coefficient of the squared term..4, by

the constant.. 3;   4(3) = 12.   We need factors of 12 that add to the

coefficient of the x term.. -7.     The factors of 12 that add to -7 are (-4)(-3)

 

Split the -7x into -4x - 3x and rewrite the equation so that it can be factored.

 

4x² -7x + 3 = 4x²-4x-3x + 3 = 4x(x-1) -3(x-1) = (4x-3)(x-1)

 

Thus we have:

 

(4x-3)(x-1) > 0

 

For this to be true either:  4x-3>0  AND  x-1>0   or   4x-3<0 AND x-1<0

 

4x-3>0     x-1>0      OR     4x-3<0       x-1<0

4x>3         x>1                 4x<3 

x>3/4  and x>1                  x<3/4  and  x<1

   This gives                          This gives

        x>1                                 x<3/4

 

Thus our domain for x is all numbers greater than 1 and all numbers less than 3/4

 

D = (-∞,3/4)U(1,∞)

Answer 2

f(x) = 1 / [√(4x² - 7x + 3)]

 

 

We know that the denominator cannot be equal to zero.  The value under the square-root cannot be negative.  So our domain can be found from the inequality

 

4x² - 7x + 3 > 0

 

(4x - 3)(x - 1) = 0

 

 

x = 3/4         and         x = 1

 

are our zeros.  We use test points using these zeros to determine which x values make the expression positive.   Use

x=0.5  , x = 0.8   and   x = 2.

 

 

(4(0.5) - 3)(0.5 - 1) = (-1)(-0.5) = positive number

 

(4(0.8) - 3)(0.8 - 1) = (0.2)(-0.2) = negative number

 

(4(2) - 3)(2 - 1) = (5)(1) = positive number

 

 

Your domain is   (-∞, 3/4)∪(1, ∞).

 

 

 

 

Answer 3

Hi Erin.

 

If you factor the denominator you get (4x-3)(x-1)

The denominator can't be zero and it also can't be negative under the square root symbol.

 

Since it can't be zero, set 4x-3 = 0 and x-1 = 0. Then x = 3/4 and x = 1 are both excluded from the domain.

 

Since it can't be negative (4x-3)(x-1) must  be > 0 and we can break the number line into intervals and choose test points.

 

x < 3/4                                        3/4<x<1                         x >1

choose 0                                     choose 7/8                     choose 2

(0-3)(0-1)                              (4(7/8)-3)((7/8)-1)              (8-3)(2-1)

(-3)(-1)                                    (1/4)(-1/8)                         (5)(1)

=3                                             = -1/32                              = 5

which is greater than zero        less than zero                  greater than 0

so this interval is good                     so out                            so good

 

(–∞, 3/4) ∪ (1, ∞)  -use parenthesis because 3/4 and 1 are not included and because you never reach + or - infinity.

 

If you see this in the next few minutes let me know if it is correct or not and if so, if you need anything else clarified.