Find the domain and range of x+1/x-2.

Answer 1

f(x) = (x + 1)/(x - 2), where x is a real number.

Domain is the set of all possible input values for which a function is defined.

Because division by 0 is undefined, x cannot be 2.

Then Domain:  x is any real number, except x ≠ 2.  This can be noted in

Set notation Domain: { x ∈ R | x ≠ 2 } OR

Interval Notation Domain: (-∞, 2) ∪ (2, +∞)

Range is the set of all possible output values of f(x) that the function produces.

x + 1 x - 2 + 2 + 1 (x - 2) + 3 3

f(x) = ------- = ----------------- = ---------------- = 1 + --------, this is a convenient form to figure out the Range

x - 2 x - 2 x - 2 x - 2

If x → 2, then (x - 2) → 0, then 3 / (x - 2) → ± ∞, depending on how x tends to 2:

If x → 2 from the left, that is x < 2, then 3 / (x - 2) → -∞, then f(x) = (x + 1) / (x - 2) → -∞

If x → 2 from the right, that is x > 2, then 3 / (x - 2) → +∞, then f(x) = (x + 1) / (x - 2) → +∞

Now, can f(x) produce any value, between -∞ and +∞?

If x takes any value, except 2,

then (x - 2) produces any value, except 0,

then 3 / (x - 2) produces any value, except 0, (because the numerator is constant 3, never 0)

then 1 + 3 / (x - 2) produces any value, except 1.

Then Range:  f(x) is any real number, except f(x) ≠ 1. This can be noted in

Set notation Range: { f(x) ∈ R | f(x) ≠ 1 } OR

Interval notation Range: (-∞, 1) ∪ (1, +∞)

Answer 2

Finding the Domain (Assuming f(x)= (x+1)/(x-2))

When finding the domain of a rational function, the first question to ask ourselves is 'what number makes the denominator zero or the function undefined?'. Once this question is answered, we can now proceed to exclude that number from the set of real numbers, because that number is what would make that equation undefined which we don't want.

Illustration

f(x)= (x+1)/(x-2)

x-2=0, x = 2

∴ 2 is to be excluded

So the domain would be (-∞, 2) U (2, ∞)

Finding the Range (Assuming f(x)= (x+1)/(x-2))

To find the range of a rational function, the easiest way is to find the horizontal asymptote. To do this, we simply divide the top and bottom by the highest power of x on the denominator.

Illustration

f(x)= (x/x + 1/x) / (x/x - 2/x)

1/x and 2/x become zero and we're left with x/x on the top which is 1, divided by x/x on the bottom which is also 1 and that gives us a horizontal asymptote of 1.

Now that we have the horizontal asymptote, we will exclude it from the set of real numbers, so the range is going to be (-∞, 1) U (1, ∞).

Answer 3

I am going to assume the function is f(x) = x + 1/(x-2).

Domain:

We know that the domain is the set of all possible x-values that we are allowed to put into the function. The first part is x which can be substituted by any number. However, in the second part, we have a fraction and a fraction cannot have 0 in its denominator because then it would be undefined. So, to find out the values of x for which the fraction is undefined, we can set the denominator to 0:

x - 2 = 0

x = 2

Therefore, we may use every real number except 2:

Domain = (-∞, 2) ∪ (2, ∞)

Note that the domain has been written in interval notation, where the parentheses around 2 mean that 2 is not included.

Range:

The range is the set of all possible output values, or y-values. Start by replacing f(x) with y:

y = x + 1/(x-2)

We want to determine which y-values this equation can produce. To do that, we can rearrange the equation into a quadratic equation in c.

y - x = 1/(x - 2)

(y - x)(x - 2) = 1

yx - 2y - x² + 2x = 1

x² - (y + 2)x + (2y + 1) = 0

This equation must have at least one real x-value for a particular y-value to be part of the range. For a quadratic equation to have a real solution, its discriminant must satisfy:

b² - 4ac ≥ 0

We know that in our quadratic, a = 1, b = -(y + 2), c = 2y + 1. Let's substitute these into the discriminant and simplify:

[-(y + 2)]² - 4(1)(2y + 1) ≥ 0

(y + 2)² - 4(2y + 1) ≥ 0

y² + 4y + 4 - 8y - 4 ≥ 0

y² - 4y ≥ 0

y(y - 4) ≥ 0

y ≤ 0 or y ≥ 4

Therefore, the function can produce outputs that are 0 or less, and outputs that are 4 or greater. It cannot produce any y-value between 0 and 4.

Hence, the range is:

Range: (-∞, 0] ∪ [4, ∞)

Note that the range has been written in interval notation, where the brackets around 0 and 4 mean that those values are included.

Answer 4

I am assuming that the function is x + 1/(x - 2).

The domain is the set of all values that can be used as input without leading to a singularity. In this case, x cannot be equal to 2, so the domain is all real numbers except for 2. This is expressed as the union (-∞, 2) ∪ (2, ∞).

Knowing the domain, can you reason out the range?