For what values of x is #f(x)= 4x^3-12x^2 # concave or convex?

1 Answer
Jan 20, 2016

Concave on #(-oo,1)#; convex on #(1,+oo)#

Explanation:

The convexity and concavity of a function and determined by the sign of the second derivative.

  • If #f''(a)>0#, then #f(x)# is convex at #x=a#.
  • If #f''(a)<0#, then #f(x)# is concave at #x=a#.

First, find the second derivative.

#f(x)=4x^3-12x^2#
#f'(x)=12x^2-24x#
#f''(x)=24x-24#

The second derivative could change signs whenever it is equal to #0#. Find that point by setting the second derivative equal to #0#.

#24x-24=0#
#x=1#

The convexity/concavity could shift only at this point. Thus, from here, we can determine on which intervals the function will be uninterruptedly convex or concave.

Use test points around #x=1#:

When #mathbf(x<1)#:

#f''(0)=-24#

Since this is #<0#, the function is concave on the interval #(-oo,1)#.

When #mathbf(x>1)#:

#f''(2)=24#

Since this is #>0#, the function is convex on the interval #(1,+oo)#.

Always consult a graph of the original function when possible:

graph{4x^3-12x^2 [-2 5, -19.9, 5.77]}

The concavity does seem to shift around the point #x=1#. When #x<1#, the graph points downward, in the #nn# shape characteristic of concavity. When #x>1#, the graph points upward in the #uu# shape characteristic of convexity.