# `f(x) = -x^4 + 4x^3 + 8x^2` Find all relative extrema, use the second derivative test where applicable. Given: `f(x)=-x^4+4x^3+8x^2`

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

`f'(x)=-4x^3+12x^2+16x=0`

`-4x(x^2-3x-4)=0`

`-4x(x-4)(x+1)=0`

`x=0, x=4,x=-1`

If f'(x)>0 the function is increasing over an interval.

If f'(x)>0 the function is decreasing over an interval.

Choose a value for x that is...

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Given: `f(x)=-x^4+4x^3+8x^2`

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

`f'(x)=-4x^3+12x^2+16x=0`

`-4x(x^2-3x-4)=0`

`-4x(x-4)(x+1)=0`

`x=0, x=4,x=-1`

If f'(x)>0 the function is increasing over an interval.

If f'(x)>0 the function is decreasing over an interval.

Choose a value for x that is less than -1.

f'(-2)=48 Since f'(2)>0 the function increases in the interval (-`oo,-1).`

Choose a value for x that is between -1 and 0.

f'(-.5)=-4.5 Since f'(-.5)<0 the function decreases in the interval (-1, 0).

Choose a value for x that is between 0 and 4.

f'(1)=24 Since f'(1)>0 the function increases in the interval (0, 4).

Choose a value for x that is greater than 4.

f'(5)=-120 Since f'(5)<0 the function decreases in the interval (4, `oo).`

Because the function changes direction from increasing to decreasing a relative maximum will exist at x=-1 and at x=4. The relative maximum points are

(-1, 3) and (4, 128).

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

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