# `f(x) = 36x + 3x^2 - 2x^3` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use...

`f(x) = 36x + 3x^2 - 2x^3` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the information from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one.

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Given: `f(x)=36x+3x^2-2x^3`

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

`f'(x)=36+6x-6x^2=0`

`f'(x)=6+x-x^2=0`

`f'(x)=x^2-x-6=0`

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

The critical values are x=3 and x=-2

Part a)

If f'(x)>0 then the function is increasing on the interval.

If f'(x)<0 then the function is decreasing on the interval.

Select an x value in the interval (- `oo` ,-2).

Since f'(-3)<0 the function is decreasing in the interval (-`oo` ,-2).

Select an x value in the interval (-2,3).

Since f(0)>0 the function is increasing in the interval (-2, 3 ).

Select an x value in the interval (3,`oo).`

Since f(4)<0 the function is decreasing in the interval (3,`oo` ).

Part b)

Because the function changed direction from decreasing to increasing at x=-2 a local minimum exists. The local minimum occurs at the point (-2,-44).

Because the function changed direction from increasing to decreasing at x=3 a local maximum exists. The local maximum occurs at the point (3, 81).

Part c)

Find the critical values for the second derivative.

`f''(x)=6-12x=0`

`6=12x`

`x=1/2`

The critical value is x=1/2.

If f''(x)>0 then the graph is concave up in the interval.

If f"(x)>0 then the graph is concave down in the interval.

If f'(x)=0 then an inflection point will exist.

Select an x value in the interval (-`oo` , 1/2).

Since f"(0)>0 the graph is concave up in the interval (-`oo` ,1/2).

Select an x value in the interval (1/2, `oo` ).

Since f"(1)<0 the graph is concave down in the interval (1/2, `oo` ).

Since f"(1/2)=0 there is an inflection point at x=1/2. The inflection point occurs at the coordinate (1/2, 18.5).