Answer to Question #290749 in Calculus for Neha

Question

Trace the curve y=x³-x

Answer 1

1) We calculate the roots:

$x^3 -x = 0, \;\; x(x^2-1) = 0, \\ x = 0 \;\;\text{or} \;\; x = 1 \;\;\text{or} \;\; x = -1 .$

2) Next, we shall determine maxima and minima:

$(x^3-x)' = 3x^2 - 1,\\ 3x^2 - 1 = 0,\\ x = \pm \dfrac{1}{\sqrt{3}}.$

At $x = -\dfrac{1}{\sqrt{3}}$ the function $(x^3-x)'$ changes sign from + to -, so it is the point of maximum. Another point is the point of minimum.

At $x = -\dfrac{1}{\sqrt{3}}$ the function $x^3-x$ obtains value of $\dfrac{2}{3\sqrt{3}},$ at $x = +\dfrac{1}{\sqrt{3}}$ the function $-\dfrac{2}{3\sqrt{3}}.$

3) Next, we'll determine if the function is odd or even.

$f(x) = x^3 - x, \\ f(-x) = (-x)^3 - (-x) = -x^3 + x = -f(x),$ so the function is odd.