For f(x) = -x^3 - 6x^2 - 2 determine all the critical points, test each interval.  Also, use the first and seconds derivative test to determine where the grapg is increasing or decreasing and...

For f(x) = -x^3 - 6x^2 - 2 determine all the critical points, test each interval.

 

Also, use the first and seconds derivative test to determine where the grapg is increasing or decreasing and which critcical points are max and mins.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given `f(x)=-x^3-6x^2-2` .

(1) The critical points occur where `f'(x)=0` or `f'(x)` fails to exist. This is a polynomial, hence continuous everywhere and infinitely differentiable so we need only find where `f'(x)=0` .

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

`f'(x)=0=>-3x(x+4)=0=>x=0` or `x=-4`

Thus the critical points are x=0 and x=-4.

(2) We test a point in each of the following intervals: `(-oo,-4),(-4,0),(0,oo)`

On `(-oo,-4),f'(x)<0` So the function decreases, check `f'(-5)`

On `(-4,0),f'(x)>0` So the function increases, check `f'(-2)`

On `(0,oo),f'(x)<0` So the function decreases, check `f'(1)`

(3) The second derivative describes the concavity of the function:

`f''(x)=-6x-12` .`f''(x)=0=>x=-2`

Testing the two intervals we find:

`f''(x)>0` on `(-oo,-2)` so the function is concave up.

`f''(x)<0` on `(-2,oo)` so the function is concave down.

(4) Since `f''(x)>0` at x=-4, x=-4 is a relative minimum.

Since `f''(x)<0` at x=0, x=0 is a relative maximum.

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