1. ## Concave/inflection Points

Determine where the graph of the given function is concave upward and concave downward. Find the coordinates of all inflection points.

f(x)=x^3+3x^2+x+1

up is [-1, infinity)
down is (- infinity, 2]
inflection points (-1,2)

I know you have to find the first and second derivative which is

f'(x)=3x^2+6x+1
f''(x)=6(x+1)

I just don't know where to go from there..Thanks!

2. Originally Posted by lisa1984wilson
Determine where the graph of the given function is concave upward and concave downward. Find the coordinates of all inflection points.

f(x)=x^3+3x^2+x+1

up is [-1, infinity)
down is (- infinity, 2]
inflection points (-1,2)

I know you have to find the first and second derivative which is

f'(x)=3x^2+6x+1
f''(x)=6(x+1)

I just don't know where to go from there..Thanks!

well youre right! Weve gotta find the second derivative. Lettss do it>

[tex]f'(x)=3x^2+6x+1[tex], and $\displaystyle f''(x)=6(x+1)$

good good good. so then the inflection point

$\displaystyle f''(x)=6(x+1)=0\Longleftrightarrow{x=-1}$

good good Now plugging back in the original formula gives us (-1,2).

good good Concavity

• x = a is a locally a minimum if f"(a) > 0,
• x = a is a locally a maximum if f"(a) < 0; and
Using a little noodle power, you can do the rest..........

You can DO IT!!!!!!!!