How do you find the inflection points of the graph of the function #f(x) = x^3 - 3x^2 + 3x#?

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Answer 1 · 2015-04-06T18:08:02

Find the points on the graph where the concavity changes.

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

So, #f'(x) = 3x^2 - 6x + 3#.

And #f''(x) = 6x - 6#.

#f''(x)=0# for #x=1#. Testing on each side of #1# we find that

#f''(x) < 0# (so the graph of #f# is concave down) for #x<1#
#f''(x) > 0# (so the graph of #f# is concave up) for #x>1#

At #x=1#, we have #y=f(1)=3-3+1=1#.

The inflection point is #(1, 1)#.

Answer 2 · 2015-04-06T18:17:24

You must study your second derivative:
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