Question

How do you find the intervals of increasing and decreasing using the first derivative given `y=(x-1)^2(x+3)`?

Answer 1

The function `y=(x-1)^2(x+3)` is rising in the interval `(-oo,-5/3)` and then it starts declining in the interval `(-5/3,1)` and then again rising in the interval `(1, oo)`

Explanation:

As `y=(x-1)^2(x+3)`

`(dy)/(dx)=2(x-1)xx(x+3)+(x-1)^2`

= `2(x^2+3x-x-3)+x^2-2x+1`

= `3x^2+2x-5`

= `3x^2-3x+5x-5`

= `3x(x-1)+5(x-1)`

= `(3x+5)(x-1)` whose zeros are `-5/3` and `1`

Now using Sign Chart

`color(white)(XXXXXXXXXXX)-5/3color(white)(XXXXX)1`

`(x-1)color(white)(XXX)-ive color(white)(XXXX)-ive color(white)(XXXX)+ive`

`(3x+5)color(white)(XXX)-ive color(white)(XXXX)+ive color(white)(XXXX)+ive`

`(3x+5)(x-1)color(white)()+ive color(white)(XXX)-ive color(white)(XXXX)+ive`

Hence, the function `y=(x-1)^2(x+3)` is rising in the interval `(-oo,-5/3)` and then it starts declining in the interval `(-5/3,1)` and then again rising in the interval `(1, oo)`

`-5/3` amd `1` are local maxima and minima.

graph{(x-1)^2(x+3) [-5, 5, -10, 10]}