# Determine where the function is concave upward and downward Determine where the function is concave upward and downward, and list all the x- value(s) of the point(s) of inflection:...

Determine where the function is concave upward and downward

Determine where the function is concave upward and downward, and list all the x- value(s) of the point(s) of inflection: f(x)=3x^3+(1)⋅x^2+1x−9. (Do not concatenate your answers. i.e., if the intervals where the function is concave downward are, e.g., (0,1) and (1,3), don't write (0,3); list the intervals separately. For ∞, type inf , and for −∞, type - inf .) The function is concave downward on the interval(s) . The function is concave upward on the interval(s) . The x- value(s) of the point(s) of inflection is (are) . (If there are none, then enter none .)]

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Find the inflection points (if any) and the intervals where the function is concave up and concave down for `f(x)=3x^3+x^2+x-9` :

A function is concave up on an interval if the second derivative is positive on the interval; concave down if the second derivative is negative.

`f(x)=3x^3+x^2+x-9`

`f'(x)=9x^2+2x+1`

`f''(x)=18x+2`

`f''(x)=0 ==>x=-1/9`

We try test points on the intervals `(-oo,-2/9),(-2/9,oo)` :

`f''(-1)=-16<0` so the function is concave down on `(-oo,-2/9)`

`f''(0)=2>0` so the function is concave up on `(-2/9,oo)`

Since the concavity changes from down to up at `x=-2/9` and the second derivative exists there` `, `(-2/9,-2237/243)` is the only inflection point.

The graph:

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