# How do you sketch the curve y=x^5-x by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Feb 7, 2017

See explanation and Socratic graph.

#### Explanation:

Sufficient data for studying the shape and extent of the graph:

x-intercepts $\left(y = 0\right) : 0 , \pm 1$

y-intercept ( x = 0 ) : 0.

$y ' = 5 {x}^{4} - 1 = 0$, at $x = \pm \frac{1}{\sqrt{\sqrt{5}}} = \pm 0.66874$, 4early

$y ' ' = 20 {x}^{3} = 0$, at x = 0.

The turning points are $\left(\pm \frac{1}{\sqrt{\sqrt{5}}} , \pm \frac{\frac{4}{5}}{\sqrt{\sqrt{5}}}\right)$, giving

local ( y'' < 0 ) maximum $\frac{\frac{4}{5}}{\sqrt{\sqrt{5}}} = 0.535$ and

local ( y'' > 0 ) minimum -(4/5)/sqrtsqrt5=-0.535#, nearly.

See the first graph for these features.

At x = 0, y'''=y'''=0 and y'''''= 120 $\ne 0$.

See the tangent crossing the curve at POI, in the not-to-scale

second ad hoc ( for the purpose ) graph.

As $\to \pm \infty , y = {x}^{5} \left(1 - \frac{1}{x} ^ 4\right) \to \pm \infty$.

graph{(x^5-x-y)((x-.67)^2+(y+.535)^2-.002)((x+.67)^2+(y-.535)^2-.002)=0 [-5, 5, -2.5, 2.5]}

graph{(x^5-x-y)(y+x)=0 [-1, 1, -2.5, 2.5]}

Tangent at the POI (0, 0) crosses the curve.