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

Apr 6, 2017

$\left\{y | y \ge 1\right\}$ or y in [1, oo[

Explanation:

The function can be rewritten as $f \left(x\right) = {x}^{2} + 2 x + 1 + 1 = {\left(x + 1\right)}^{2} + 1$. Notice that (x+1)^2≥0 for all $x \in \mathbb{R}$. Thus, the values $f \left(x\right)$ can be is all values greater than or equal to $1$.

Thus, the range is $\left\{y | y \ge 1\right\}$ or y in [1, oo[.

Apr 6, 2017

Range:
$1 \le y < \infty$

$y \in \left[1 , \infty\right)$

Explanation:

Since this is an upward facing parabola, we know that the upper bound of the range is $+ \infty$

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 $f ' \left(x\right)$ is equal to zero, and find the corresponding $y$ value.

color(blue)("Method 1: Vertex form"
$f \left(x\right) = {x}^{2} + 2 x + 2$

$f \left(x\right) = {\left(x + 1\right)}^{2} + 1$

Vertex is at $\left(- 1 , 1\right)$, so lowest $y$ value is $1$.

color(red)("Method 2: Set derivative equal to zero"
$f \left(x\right) = {x}^{2} + 2 x + 2$

$f ' \left(x\right) = 2 x + 2$

$0 = 2 x + 2$

$x = - 1$

$f \left(- 1\right) = {\left(- 1\right)}^{2} + 2 \left(- 1\right) + 2 = 1$
The minimum is at $x = - 1$, and $f \left(- 1\right) = 1$ so $y = 1$ is the lowest y value.