Let f(x)=x^3−6x^2+4x−7. Find the open intervals on which f is concave up (down). Then determine the x-coordinates of all inflection points of f.
1. f is concave up on the intervals
2. f is concave down on the intervals
3. The inflection points occur at x =

Question

Let f(x)=x^3−6x^2+4x−7. Find the open intervals on which f is concave up (down). Then determine the x-coordinates of all inflection points of f.
1.  	f is concave up on the intervals	
2.  	f is concave down on the intervals	
3.  	The inflection points occur at x =

anonymous · Answer

$$ f(x)=x^3−6x^2+4x−7$$ did you take the derivative?

anonymous · Answer

yeah

anonymous · Answer

then you have to take the derivative again

anonymous · Answer

by which i mean you need the second derivative
it will be a line (degree 1)

anonymous · Answer

what do you get for  $f''(x)$ ?

anonymous · Answer

okay i took the second derivative how im suppose to find the concavity

anonymous · Answer

if the second derivative is positive, then the function is "concave up"

anonymous · Answer

and if the second derivative is negative, the function is concave down
the inflexion point is where the second derivative is 0

anonymous · Answer

but i think i have to plug in critical points how do i solve for those

anonymous · Answer

nah no need for critical points 
it just asks "concave up" "concave down" "inflexion"

anonymous · Answer

what did you get for the second derivative?

anonymous · Answer

6x-12 .... but then how would i know hat my intervals are

anonymous · Answer

or actually "inflection" i cant spell too good

anonymous · Answer

if you want to know over what interval the function is concave up, solve 
$$6x-12>0$$ which you can almost do in your head

anonymous · Answer

so 2 would be that its positive and that means its concave up

anonymous · Answer

ok lets go slow

anonymous · Answer

you are looking for an interval, not a number, an interval over which the function is concave up
the second derivative is $6x-12$ and the function will be concave up if $6x-12>0$ i.e. it is concave up over the interval for which $6x-12>0$ 
if $x>2$ then $6x-12>0$ and so the function is concave up on the interval 
$$(2,\infty)$$

anonymous · Answer

the fact that 2 is positive has nothing to do with the function being concave up, it is just the endpoint of the interval over which $6x-12>0$

anonymous · Answer

and of course if $6x-12<0$ that means $x<2$ so the function is concave down on $$(-\infty,2)$$

anonymous · Answer

if you think about what a cubic function with positive leading coefficient looks like, this makes perfect sense |dw:1366942287621:dw|

anonymous · Answer

my point of inflection would be at 2 right

anonymous · Answer

yup

anonymous · Answer

thank you

anonymous · Answer

yw