Is #f(x)=(x-3)(x+2)(x-4)^2# concave or convex at #x=-1#?

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Answer 1 · 2017-03-07T14:19:27

The function is convex at #x=-1#

We need

#(uvw)'=u'vw+v'uw+w'uv#

We must calculate the first and second derivatives

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

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

#=(x-4){(x+2)(x-4)+(x-3)(x-4)+2(x-3)(x+2)}#

#=(x-4){x^2-2x-8+x^2-7x+12+2x^2-2x-12}#

#=(x-4)(4x^2-11x-8)#

#f''(x)=(4x^2-11x-8)+(x-4)(8x-11)#

#=4x^2-11x-8+8x^2-43x+44#

#=12x^2-54x+36#

Therefore,

#f''(-1)=12+54+36=102#

As, #f''(-1)>0#, we conclude that the function is convex