Question

What is the range of the function `f(x)=x^2+2x+2`?

Answer 1

`{y|yge1}` or `y in [1, oo[`

Explanation:

The function can be rewritten as `f(x)=x^2+2x+1+1=(x+1)^2+1`. Notice that `(x+1)^2≥0` for all `x in RR`. Thus, the values `f(x)` can be is all values greater than or equal to `1`.

Thus, the range is `{y|yge1}` or `y in [1, oo[`.

Answer 2

Range:
`1 <= y < oo`

`y in [1,oo)`

Explanation:

Since this is an upward facing parabola, we know that the upper bound of the range is `+oo`

Find the minimum. We can do this by either using our knowledge of parabolas and quadratic form to complete the square to find the vertex, and the y value is the minimum; OR we could use calculus and find where `f'(x)` is equal to zero, and find the corresponding `y` value.

`color(blue)("Method 1: Vertex form"`
`f(x)=x^2+2x+2`

`f(x)=(x+1)^2+1`

Vertex is at `(-1,1)`, so lowest `y` value is `1`.

`color(red)("Method 2: Set derivative equal to zero"`
`f(x)=x^2+2x+2`

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

`0=2x+2`

`x=-1`

`f(-1)=(-1)^2+2(-1)+2=1`
The minimum is at `x=-1`, and `f(-1)=1` so `y= 1` is the lowest y value.