For what values of x is #f(x)=(x+2)(x-3)(x+1)# concave or convex?

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Answer 1 · 2017-04-13T12:37:42

The function is concave for #x in (-oo,0)#
The function is convex for #x in (0,+oo)#

We need

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

We must calculate the second derivative and determine the sign.

#f(x)=(x+2)(x-3)(x+1)#

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

#=x^2-2x-3+x^2+3x+2+x^2-x-6#

#=3x^2-7#

#f''(x)=6x#

Therefore,

#f''(x)=0# when #x=0#

This is the point of inflexion.

We can build a chart

#color(white)(aaaa)##Interval##color(white)(aaaaaa)##(-oo,0)##color(white)(aaaaaa)##(0,+oo)#

#color(white)(aaaa)##sign f''(x)##color(white)(aaaaaaa)##-##color(white)(aaaaaaaaaaa)##+#

#color(white)(aaaa)## f(x)##color(white)(aaaaaaaaaaaaa)##nn##color(white)(aaaaaaaaaaa)##uu#