For this rational function f, f(x) 3x-6/x-1. determine where possible

Question

1. For this rational function f, f(x) 3x-6/x-1. determine where possible

 

i. The domain of f

 

ii. x and y-intercept of f in coordinate form.

 

iii. The vertical asymptote(s) and oblique asymptote of the graph of f.

 

iv. Is f a one on one function? Explain

 

v. Is f an even or an odd function? Explain

 

vi.Use the graph of f to solve the inequalities f(x) greater than equal 0 and f(x) less than equal to 0

 

vii. Find the domain of g and h, given that g(x)= square root of f(x) and h(x)= log f(x)

Answer 1

Hi Vivian.

 

i) To find the domain you set the denominator equal to zero and solve. x - 1 = 0 so x = 1. Since x = 1 will make the denominator zero, it is excluded from the domain and so the domain is "All real numbers, x ≠ 0". There is no other number that would need to be excluded from the domain.

 

ii) To find the x-intercept, let y = 0 and solve for x, which is equivalent to setting the numerator equal to 0. So in this case 3x - 6 = 0 and x = 2. So the x-intercept is (2, 0). To find the y-intercept, let x = 0 and solve for y. So y = (3(0)-6)/(0-1) = -6/-1 = 6. So the y-intercept is (0,6)

 

iii) The vertical asymptote is the vertical line through the value for x which is excluded from the domain. So the vertical asymptote is x = 1. (Because x can never equal 1 the graph will never have a point anywhere on that line.)

 

This graph has a horizontal asymptote at y = 3. A graph can only have a horizontal or oblique asymptote but not both. 

The horizontal asymptote is at y = 3 because the degree of the numerator is the same as the degree of the denominator. When that happens you divide the coefficients of the two leading terms (when written in descending order) to find the horizontal asymptote. 3x/x = 3.

 

iv) To see if it's a one-to-one function, see if substituting in two different variables leads to the conclusion that the variables are equal. So 3a-6/a-1 and 3b-6/b-1. To test it we set them equal and simplify. 3a-6/a-1 = 3b-6/b-1. In the first step we cross multiply so (3a-6)(b-1) = (3b-6)((a-1). Then foil to get 3ab-6b-3a+6=3ab-6a-3b+6. Subtract the 3ab and the +6 from both sides to get -6b-3a = -6a-3b. Add 3b to both sides: -3b-3a=-6a. Add 3a to both sides:

-3b=-3a. Divide both sides by -3: b=a. So yes, the function is 1-1. A graph of the function also shows that it passes the horizontal line test, another test for 1-1.

 

v) To test for even, substitute -x for x and see if the function is equivalent. So (3(-x)-6)/((-x)-1) = (-3x-6)/(-x-1) = 3x+6/x+1 which is not the same function so it is not even. If you graph it you can also see that it is not symmetric about the y-axis. For odd see if the function is the opposite of the original which would be -3x+6/-x+1 but we got 3x+6/x+1 - so the function is neither even nor odd. 

 

vi) For f(x) ≥ 0 and f(x) ≤ 0, break the graph into sections based on the x-intercept (2,0) and the vertical asymptote (x=1). So the sections are x < 1, 1 < x < 2, and x > 2. Then pick a test point in each section, so 0, 1.5 and 3, and sub each of them into the equation. f(0) = 6 (we already calculated that above); f(1.5) = 3(1.5)-6/(1.5)-1 = -1.5/.5 = -3; f(3) = 3(3)-6/3-1 = 3/2. Since 6 and 3/2 are positive, f(x) ≥ 0 for x < 1 and x > 2. Since -3 is negative f(x) ≤ 0 for 1 < x < 2.

 

vii) The domain of g would exclude x = 1 since that would still make the denominator zero. Also excluded would be negatives under the radical, which would be the same area that gave a negative in part vi) so the domain would be (-∞, 1) ∪ [2, ∞). For h, logs can't take negative arguments, so I believe the domain would be the same as for g.

 

Hope this helps you to understand these concepts better. Also, I checked my work, but apologize if I made a mistake anywhere in the solution!

 

Sincerely,

Diana R.