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

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Answer 1 · 2017-03-19T11:20:57

$f(x)$ has a real zero at $x~=- 1.148698$ and an inflection point at $(0,2)$
$f(x)$ is defined $forall x in RR$

$f(x) = x^5+2$

Consider: $f(x) =0 -> x^5 = -2$

$:. x = root5(-2) = -1.148698$ for $x in R$

Now consider, $f(0) = 0^5 + 2= 2$

$f'(x) = 5x^4$
$f''(x) = 20x^3$

Hence $f''(0) = 0$ $-> f(x)$ has an infection point at $(0, 2)$

Finally realise , the domain and range of $f(x)$ is $(-oo, +oo)$

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

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

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