1. ## inflection points

Let f(x)= -x^4-7x^3+8x+3.
(A) Use interval notation to indicate where is concave up.
Note: Use 'INF' for , '-INF' for , and use 'U' for the union symbol.
Concave up:

(B) Use interval notation to indicate where f(x) is concave down.
Concave down:
(C) List the values of all the inflection points of . If there are no inflection points, enter 'NONE'.
values of inflection points =

2. Originally Posted by abcdef
Let f(x)= -x^4-7x^3+8x+3.
(A) Use interval notation to indicate where is concave up.
Note: Use 'INF' for , '-INF' for , and use 'U' for the union symbol.
Concave up:

(B) Use interval notation to indicate where f(x) is concave down.
Concave down:
(C) List the values of all the inflection points of . If there are no inflection points, enter 'NONE'.
values of inflection points =
$\displaystyle f(x) = -x^4 - 7x^3 + 8x + 3$

$\displaystyle f'(x) = -4x^3 - 21x^2 + 8$

$\displaystyle f''(x) = -12x^2 - 42x$

solving for $\displaystyle f''(x) = 0$ yields $\displaystyle x=0$ and $\displaystyle x=-\frac{7}{2}$
so, the intervals for testing are $\displaystyle (-\infty ,-\frac{7}{2}), (-\frac{7}{2}, 0),(0,+\infty)$

if x is in $\displaystyle (-\infty ,-\frac{7}{2})$, then $\displaystyle f''(x) < 0$

if x is in $\displaystyle (-\frac{7}{2}, 0)$, then $\displaystyle f''(x) > 0$

if x is in $\displaystyle (0,+\infty)$, then $\displaystyle f''(x) < 0$

given that, when is the function concave up or down? Ü