Answer to Question #100431 in Calculus for Harsh

Question

Trace the curce
Y=(x-2)(x+1)^2

Answer 1

1 The points, where the function value Y is equal zero: $x=2$ and $x=-1$ .

2 Consider intervals.

When $x > 2$ the function value Y is positive.

When $-1< x < 2$ the function value Y is negative.

When $x < -1$ the function value Y is also negative.

3 Let's transform the function expression:

$Y = (x-2)(x+1)^2 = (x-2)(x^2 +2x +1) =$

$(x^3 + 2 x^2 +x) - 2x^2 -4x - 2= x^3 -3x - 2$

Now we find the first derivative of the function:

$Y' = ( (x-2)(x+1)^2 )' = (x^3 -3x - 2)' = 3 x^2 - 3$

So, the function is smooth and its derivative $Y'$ is equal zero at poins $x=1$ and $x=-1$ .

Also, $Y'<0$ when $-1 < x < 1$ (on this interval the function value is decreasing), and $Y'>0$ when $x<-1$ and $x>1$ (in these intervals the function value is growing). So, it changes its sign from negative to positive at $x = 1$ and from positive to negative at $x = -1$. That is why the function has extrema at $x=-1$ (maximum) and at $x =1$ (minimum).