Answer to Question #94476 in Calculus for Nikita

Question

Trace the curve y^2=(x+1)(x-1)^2 by showing all properties

Answer 1

The curve is $y^2=(x+1)(x-1)^2$

$(x+1)(x-1)^2\geq 0 \Rightarrow (x+1)\geq 0 \Rightarrow x\geq -1 .$

$y=\pm \sqrt{(x+1)(x-1)^2}$

The graph of the curve is symmetric about the x-axis.

Graph of function $y= \sqrt{(x+1)(x-1)^2}$ :

$y(-1)= 0, y(-1)= 0.$

$y'(x)= \frac{3x^2+2x-1}{ \sqrt{(x+1)(x-1)^2}}$

$y'(-1/3)= 0 ;$

$y'(1)$ - not exist.

The function increases for x in the interval $[-1;-\frac{1}{3}] and [1;\infty], (y'>0)$

The function decreasing for x in the interval $[-\frac{1}{3};1], (y'<0)$

Let us try to find where a function is increasing or decreasing

Graph of the curve :