# How do you graph f(x)=x^5+2 using zeros and end behavior?

Mar 19, 2017

$f \left(x\right)$ has a real zero at $x \cong - 1.148698$ and an inflection point at $\left(0 , 2\right)$
$f \left(x\right)$ is defined $\forall x \in \mathbb{R}$

#### Explanation:

$f \left(x\right) = {x}^{5} + 2$

Consider: $f \left(x\right) = 0 \to {x}^{5} = - 2$

$\therefore x = \sqrt[5]{- 2} = - 1.148698$ for $x \in R$

Now consider, $f \left(0\right) = {0}^{5} + 2 = 2$

$f ' \left(x\right) = 5 {x}^{4}$
$f ' ' \left(x\right) = 20 {x}^{3}$

Hence $f ' ' \left(0\right) = 0$ $\to f \left(x\right)$ has an infection point at $\left(0 , 2\right)$

Finally realise , the domain and range of $f \left(x\right)$ is $\left(- \infty , + \infty\right)$

These attributes of $f \left(x\right)$ on the real xy-plane can be seen on the graph below.

graph{x^5+2 [-10, 10, -5, 5]}