# How do you sketch the general shape of f(x)=x^5-3x^3+2x+4 using end behavior?

Jan 21, 2017

See graph and explanation. The second graph reveals turning points and points of inflexion (POI).

#### Explanation:

graph{x^5-3x^3+2x^2+4 [-39.86, 39.8, -20.43, 19.43]}

$f = {x}^{5} \left(1 - \frac{3}{x} ^ 2 + \frac{2}{x} ^ 4 + \frac{4}{x} ^ 5\right) \to \pm \infty$, as $x \to \pm \infty$.

$f ' = 3 {x}^{4} - 6 {x}^{2} + 2$, giving turning points?POI, at its

zeros x = +-1.24 and +-0.51, nearly,

$f ' ' = 12 x \left({x}^{2} - 1\right) = 0$, at $x = 0 \mathmr{and} x = \pm 1$.

$f ' ' ' = 12 \left(3 {x}^{2} - 1\right) \ne 0$, at these points.

So, the POI are at $x = 0 , \pm \frac{1}{\sqrt{3}}$.

The second graph locates the turning points and POE that could not

be located in the first graph

.graph{x^5-3x^3+2x^2+4[-1.5, 1.5, -20.43, 19.43]}