Is #f(x)=(x-3)^3-x+15# concave or convex at #x=3#?

1 Answer
Aug 14, 2017

Answer:

Neither. #x= 3# is an inflection point.

Explanation:

You have to find the second derivative to answer this question.

I would use the chain rule to differentiate #(x- 3)^3#. Let #u = x - 3# and #y = u^3#. Then #y' = 3u^2 * u' = 3u^2 * 1 = 3(x - 3)^2#

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

Now it's easy enough to expand to find the second derivative using the power rule.

#f'(x) = 3(x^2 - 6x - 9) - 1#

#f'(x) = 3x^2 - 18x - 27 - 1#

#f'(x) = 3x^2 - 18x - 28#

Now find the second derivative.

#f''(x) = 6x - 18#

We now test the sign of the second derivative at #x = 3# under the knowledge that if:

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

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

We have:

#f''(3) = 6(3) - 18 = 18 - 18 = 0#

So #x = 3# is a point of inflection, so the graph is neither concave up nor concave down at that point. Instead its changing between the two.

Hopefully this helps!