Homework Help

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

jjmgingrich's profile pic

Posted via web

dislike 2 like

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=?

1 Answer | Add Yours

embizze's profile pic

Posted (Answer #1)

dislike 0 like

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`

-----------------------------------------------------------------

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.

------------------------------------------------------------------

The graph:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes