How do you find Find the points of inflection and discuss the concavity of the function #f(x) = 2x^3-3x^2-12x+5#?

1 Answer
Apr 3, 2015

Concavity and inflection points are investigated using the second derivative and its sign.

#f(x) = 2x^3-3x^2-12x+5#

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

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

Now, #f''(x)# is continuous, so if it changes sign it must be #0# at some point (Intermediate Value Theorem).
(A discontinuous function can also change sign at a discontinuity.)

#f''(x) = 12x-6 = 0# at #x=1/2#

Left of #1/2# (say at #x=0#, we see that #f''(x)<0# So the graph of #f# is concave down to the left of #x=1/2#

If #x# is right of #1/2#, (say at #x=10#) then #f''(x)>0# so the graph of #f# is concave up to the right of #x=1/2#.

The concavity changes as we cross #x=1/2# if there is a point on the graph at #x=1/2#, (if #1/2# is in the domain of #f#) then that point is an inflection point.

#f(1/2) = 2(1/2)^3-3(1/2)^2-12(1/2)+5 = 1/4-3/4-6+5=-2/4-1=-3/2#

So the point #(1/2, 3/2)# is an inflection point.