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Let f(x)=-x^4-2x^3+6x-4. Find the open intervals on which  is concave up (down)....

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jjmgingrich | Student, Undergraduate | (Level 1) Salutatorian

Posted February 26, 2013 at 2:31 AM via web

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Let f(x)=-x^4-2x^3+6x-4. Find the open intervals on which  is concave up (down). Then determine the -coordinates of all inflection points of . It is concave up on which intervals? and concave down on which intervals?   The inflection points occur at x=?

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

Posted February 26, 2013 at 4:42 AM (Answer #1)

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The concavity of the graph of a function can be determined by the second derivative; if the function is concave up on an interval then the second derivative is positive, and a negative second derivative means the graph is concave down. Inflection points occur when the second derivative changes sign.

Given `f(x)=-x^4-2x^3+6x-4`

`f'(x)=-4x^3-6x^2+6`

`f''(x)=-12x^2-12x`

Setting the second derivative equal to zero we get:

`-12x^2-12x=0`

`-12x(x+1)=0` so x=0 or x=-1.

We check the sign of the second derivative on the following intervals:

On `(-oo,-1) f''(x)<0`

On `(-1,0) f''(x)>0`

On `(0,oo)f''(x)<0`

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The function is concave down on `(-oo,-1),(0,oo)` and concave up on (-1,0). The inflection points are at x=-1 and x=0.

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The graph:

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