How do you ise interval notation indicate where f(x) is concave up and concave down for #f(x)=x^(4)-6x^(3)#?

1 Answer
Jun 16, 2017

#f(x)# is concave up from #(-oo,0)uu(3,oo)#
#f(x)# is concave down from #(0,3)#

Explanation:

What you want to do is find the second derivative using the power rule:

#d/dxx^n=nx^(n-1)#

The first derivative is:

#d/dx=4x^3-18x^2#

The second derivative is:

#f''(x)=12x^2-36x#

What you want to do now is factor it and set it equal to zero:

#12x(x-3)=0#

#12x=0# #&# #x-3=0#

#x=0,3#

Now you make a test interval from:
#(-oo,0)uu(0,3)uu(3,oo)#

You test values from the left and right into the second derivative but not the exact values of #x#.
If you get a negative number then it means that at that interval the function is concave down and if it's positive its concave up.

If done so correctly you should get that:

#f(x)# is concave up from #(-oo,0)uu(3,oo)# and that
#f(x)# is concave down from #(0,3)#

You should also note that the points #f(0)# and #f(3)# are inflection points.

Attached below is a picture that may help you:

enter image source here

The graph may also help you:

graph{x^4-6x^3 [-147.3, 145, -125.6, 20.5]}