How do you find all critical point and determine the min, max and inflection given #f(x)=2x^3-x^2+1#?

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Answer 1 · 2017-01-11T11:26:49

#f(x)# has a local maximum for #x=0#, a local minimum for #x=1/3# and an inflection point at #x=1/6#

We can find the critical points by equating the first derivative to zero:

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

#f'(x) = 6x^2-2x =2x(3x-1)#

so the critical points are:

#x_1 = 0#
#x_2 = 1/3#

To determine if they are local extrema and to find inflection points we calculate the second derivative:

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

So we have a single inflection point at #x=1/6# where #f(x)# changes from concave down to concave up.

Based on the sign of the second derivative we can also conclude that #x_1# is a local maximum and #x_2# a local minimum.

graph{2x^3-x^2+1 [-0.3227, 0.9273, 0.7125, 1.3375]}