# What is the domain and range for F(x) = -2(x + 3)² - 5?

Sep 16, 2015

Domain: ${D}_{f} = R$
Range: ${R}_{f} = \left(- \infty , - 5\right]$

#### Explanation:

graph{-2(x+3)^2-5 [-11.62, 8.38, -13.48, -3.48]}

This is quadratic (polynomial) function so there aren't points of discontinuity and hence domain is $R$ (set of real numbers).

${\lim}_{x \to \infty} \left(- 2 {\left(x + 3\right)}^{2} - 5\right) = - 2 {\left(\infty\right)}^{2} - 5 = - 2 \cdot \infty - 5 = - \infty - 5 = - \infty$

${\lim}_{x \to - \infty} \left(- 2 {\left(x + 3\right)}^{2} - 5\right) = - 2 {\left(- \infty\right)}^{2} - 5 = - 2 \cdot \infty - 5 = - \infty - 5 = - \infty$

However, function is bounded as you can see in graph so we have to find upper bound.

$F ' \left(x\right) = - 4 \left(x + 3\right) \cdot 1 = - 4 \left(x + 3\right)$
$F ' \left({x}_{s}\right) = 0 \iff - 4 \left({x}_{s} + 3\right) = 0 \iff {x}_{s} + 3 = 0 \iff {x}_{s} = - 3$

$\forall x > {x}_{s} : F ' \left(x\right) < 0 , F \left(x\right)$ is decreasing

$\forall x < {x}_{s} : F ' \left(x\right) > 0 , F \left(x\right)$ is increasing

So, ${x}_{s}$ is maximum point and

${F}_{\max} = F \left({x}_{s}\right) = F \left(- 3\right) = - 5$

Finally:

Domain: ${D}_{f} = R$
Range: ${R}_{f} = \left(- \infty , - 5\right]$