For what values of x is #f(x)= -5x^3+x^2+4x-12 # concave or convex?

Question

1248 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-08T20:25:21

#f(x)# is convex on #(-oo,1/15)#, concave on #(1/15,+oo)#, and has a point of inflection when #x=1/15#.

#f(x)# is convex when #f''(x)>0#.
#f(x)# is concave when #f''(x)<0#.

Find #f''(x)#:

#f(x)=-5x^3+x^2+4x-12#

#f'(x)=-15x^2+2x+4#

#f''(x)=-30x+2#

The concavity could change when #f''(x)=0#. This is a possible point of inflection.

#f''(x)=0#

#-30x+2=0#

#x=1/15#

Analyze the sign surrounding the point #x=1/15#. You can plug in test points to determine the sign.

When #x<1/15#, #f''(x)>0#.

When #x>1/15#, #f''(x)<0#.

When #x=15#, #f''(x)=0#.

Thus, #f(x)# is convex on #(-oo,1/15)#, concave on #(1/15,+oo)#, and has a point of inflection when #x=1/15#.

Graph of #f(x)#:

graph{-5x^3+x^2+4x-12 [-22.47, 28.85, -20.56, 5.1]}

The concavity does seem to shift very close to #x=0#.