Is #f(x)=(x-2)^3-x^4+x# concave or convex at #x=0#?

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Answer 1 · 2017-09-14T17:47:39

take the second derivative...

#f'(x) = 3(x-2)^2 - 4x^3 + 1#

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

...evaluate #f''(x)# at #x = 0#:

= 6(-2) = -12

So this means at the slopes of the tangent lines to the original function are decreasing (as x increases) at this point.

So, I'd say that the function is "concave downward" at this point. That would make it "Convex upward".

A graph of the original function is helpful:

graph{(x-2)^3 - x^4 + x [-10, 10, -5, 5]}