How do you describe the concavity of the graph and find the points of inflection (if any) for #f(x) = x^3 - 3x + 2#?

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Answer 1 · 2015-10-18T02:56:43

The function has a minimum at >#x=1# and the curve is concave upwards.

The function has a maximum at >#x=-1# and the curve is concave downwards

Given -

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

#dy/dx=3x^2-3#

#(d^2x)/(dx^2)=6x#

#dy/dx=0 => 3x^2-3=0#

#3x^2=3#

#x^2=3/3=1#

#sqrt(x^2)=+-sqrt1#

#x=1#
#x=-1#

At >#x=1# ;

#(d^2x)/(dx^2)=6(1)=6>0#

At >#x=1 ; dy/dx=0;(d^2x)/(dx^2)>0 #

Hence the function has a minimum at >#x=1# and the curve is concave upwards.

At >#x=-1# ;

#(d^2x)/(dx^2)=6(-1)=-6<0#

At >#x=-1 ; dy/dx=0;(d^2x)/(dx^2)<0 #

Hence the function has a maximum at >#x=-1# and the curve is concave downwards .

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

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