# How do you describe the end behavior for f(x)=x^5-4x^3+x+1?

Dec 19, 2016

See explanation for end trends, turning points and points of inflexion. Illustrative graph is inserted

#### Explanation:

graph{x^5-4x^3+x+1 [-3, 3, -29.54, 10.82]} $y = f \left(x\right) = {x}^{5} - 4 {x}^{3} + x + 1$

$= {x}^{5} \left(1 - \frac{4}{x} ^ 2 + \frac{1}{x} ^ 4 + \frac{1}{x} ^ 6\right) \to \pm \infty$, as $x \to \pm \infty$.

$y ' = 5 {x}^{4} - 12 {x}^{2} + 1 = 5 \left({\left({x}^{2} - \frac{6}{5}\right)}^{2} - \frac{31}{25}\right) = 0$,

when ${x}^{2} = \frac{6 \pm \sqrt{31}}{5} \to x = {+}_{\sqrt{\frac{6 \pm \sqrt{31}}{5}}}$

$= \pm 1.521 , \pm 0.2940$, giving local extrema.

$y ' ' = 20 {x}^{3} - 24 x = 20 x \left({x}^{2} - \frac{4}{5}\right) = 0$, when $x = 0 , \pm \frac{2}{\sqrt{5}}$, giving

possible points of inflexion, if f''' is not 0.

$y ' ' ' = 12 \left(5 {x}^{2} - 2\right)$ and is not 0, when y'=0.