# `f(x) = (x - 1)^2(x+ 3)` Find the critical numbers, open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

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Given: `f(x)=(x-1)^2(x+3)`

Find the critical values by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=(x-1)^2(1)+(x+3)(2(x-1))=0`

`x^2-2x+1+(x+3)(2x-2)=0`

`x^2-2x+1+2x^2-2x+6x-6=0`

`3x^2+2x-5=0`

`(3x+5)(x-1)=0`

`x=-5/3, x=1`

The critical numbers are x=-5/3 and x=1.

If f'(x)>0 the function increases in the interval.

If f'(x)<0 the function decreases in the interval.

Choose an x value less than -5/3.

f'(-2)=3 Since f'(-2)>0 the function increases in the interval (-`oo,-5/3).`

Choose an x value between -5/3 and 1.

f'(-1)=-4 Since f'(-1)<0 the function decreases in the interval (-5/3, 1).

Choose an x value greater than 1.

f'(2)=11 Since f'(2)>0 The function increases in the interval (1, `oo).`

Because the direction of the function changed from increasing to decreasing a relative maximum will exist at x=-5/3. The relative maximum is the point

(-5/3, 9.4815).

Because the direction of the function changed from decreasing to increasing a relative minimum will exist at x=1. The relative minimum is the point (1, 0).