Is #f(x)=x(x-2)(x+3)(x-1)# concave or convex at #x=1 #?

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Answer 1 · 2018-03-29T06:30:59

#f(x)# is concave DOWNWARDS (convex?) at #x=1#.

This problem may be easier if we determine the expanded (not factored) form of #f(x)# first.

#f(x)=x(x-2)(x+3)(x-1)=x(x^2+x-6)(x-1)=x(x^3-7x+6)=x^4-7x^2+6x#

#f^'(x)=4x^3-14x+6#

and

#f^('')(x)=12x^2-14#

#f^('')(1)=12*1^2-14=-2#

Which means that #f(x)# is concave DOWNWARDS (convex?) at #x=1# because #f^('')(1)<0#

graph{x(x-2)(x+3)(x-1) [-4, 3, -26, 10]}