Question

How do you find the domain and range of `f(x)=x^2+x`?

Answer 1

Domain: `(-oo, oo)`
Range: `[-1/4, oo)`

Explanation:

Given:

`f(x) = x^2+x`

As with any polynomial, this is well defined for all values of `x`, so its domain is the whole of the Real numbers `RR`, in interval notation: `(-oo, oo)`

One way of finding the domain is to complete the square:

`x^2+x = (x+1/2)^2-1/4`

Note that:

`(x+1/2)^2 >= 0`

for any Real value of `x`, with equality when `x=-1/2`

So the minimum value of `f(x)` is:

`f(-1/2) = 0^2-1/4 = -1/4`

Since `f(x)` is continuous and unbounded, we can deduce that the range is `[-1/4, oo)`

One way of proving that goes as follows.

Let:

`y = x^2+x = (x+1/2)^2-1/4`

Add `1/4` to both sides to get:

`y+1/4 = (x+1/2)^2`

Transpose and take the square root of both sides, allowing for both positive and negative square roots to get:

`x+1/2 = +-sqrt(y+1/4)`

Subtract `1/2` from both sides to find:

`x = -1/2+-sqrt(y+1/4)`

So, provided `y in [-1/4, oo)`, there is at least one value of `x` such that `y = x^2+x`. That is: `y` is in the range of `f(x)`.