Find the holes of
f(x) = (3x^2+2x-1)/(x^2-1)

Question

Find the holes of 
f(x) = (3x^2+2x-1)/(x^2-1)

anonymous · Answer

Also, may you find the range?

calculusfunctions · Answer

Step 1: Factor the numerator and the denominator of the rational function, completely.
Step 2: Divide factors common to both numerator and denominator.
Step 3: Any factors that are completely removed from the denominator, the restriction from these factors,  are the x-coordinates of the points where the removable discontinuities (holes) are. Calling them removable discontinuities is much more elegant then calling them holes, don't you think?

$$f(x) = \frac{ 3x ^{2} + 2x - 1}{x ^{2} - 1}$$ x ≠ -1, 1

Step 1:
$$f(x) = \frac{ (3x - 1)(x + 1) }{(x - 1)(x + 1) }$$

Step 2:
$$f(x) = \frac{ 3x - 1 }{x - 1  }$$

Step 3: 
The factor of  (x + 1)  was removed from the denominator. Thus there is a removable discontinuity (hole) at x = -1.

To find the y-coordinate of the point of removable discontinuity, evaluate the simplified fraction for x = -1. Thus

$$f(x) = \frac{ 3x - 1 }{ x - 1 }$$

f(-1) = 2

∴ the removable discontinuity (hole) occurs at the point (-1, 2)


You review this while I explain range.

anonymous · Answer

how did you get 2?

calculusfunctions · Answer

$$f(x) = \frac{ 3x - 1 }{ x - 1 }$$

$$f(-1) = \frac{ 3(-1)-1 }{ -1-1 }$$

$$f(-1) = \frac{ -3-1 }{ -1- 1 }$$

$$f(-1) = \frac{ -4 }{ -2 }$$

f(-1) = 2

anonymous · Answer

Wait what??? I thought the zeroes were supposed to make the equation equal zero?

calculusfunctions · Answer

You didn't ask for zeros, you asked where the removable discontinuities are. Zeros are the x-intercepts.

calculusfunctions · Answer

The x-intercepts occur when f(x) = 0. A rational function can only equal zero if the numerator of the simplified fraction equals zero. Thus in the case of the above the zero(s) (x-int.) is at x = 1/3.

anonymous · Answer

Wish me luck on the test tomorrow

calculusfunctions · Answer

I was typing the explanation to find the range and the computer rebooted. I have to start over. Give me a few minutes.

calculusfunctions · Answer

To find the range algebraically, write the simplified equation in terms of x (solve the equation for x).

$$f(x) = \frac{ 3x -1 }{ x -1 }$$

since y = f(x)

$$y = \frac{ 3x -1 }{ x -1 }$$

Multiply both sides of the equation by (x - 1) to obtain

y(x - 1) = 3x - 1

xy - y = 3x - 1

xy - 3x = y - 1

Since we need to solve for x, factor x from the left side of the equation.

x(y - 3) = y - 1

Divide both sides by (y - 3). to obtain

$$x = \frac{ y -1 }{ y -3 }$$

The restriction here is y ≠ 3.

∴ the range of f(x) is {y ∈ ℝ | y ≠ 3}.


Please let me know if this helps since your test is tomorrow, and GOOD LUCK!!