How do you determine whether the function #f(x)=2x^3-3x^2-12x+1# is concave up or concave down and its intervals?

1 Answer
Jim H
Aug 19, 2015

Investigate the sign of the second derivative.
#f# is concave down on #(-oo, 1/2)# and concave up on #(1/2, oo)#.
#(1/2, -11/2)# is the inflection point for the graph of #f#.

Explanation:

#f(x)=2x^3-3x^2-12x+1#

#f'(x)=6x^2-6x-12#

#f''(x)= 12x - 6#

#f''(x)# is never undefined and is #0# only at #x=1/2#

So the only #x# value at which the concavity might change is #1/2#

For #x < 1/2# we find that #f''(x) < 0# and for #x > 1/2#, #f''(x) > 0#

So the graph of #f# is concave down on #(-oo, 1/2)# and concave up on #(1/2, oo)#.

#f(1/2) = 2(1/2)^3-3(1/2)^2-12(1/2)+1 = -11/2#

#(1/2, -11/2)# is the only inflection point for the graph of #f#.

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