How do you describe the concavity of the graph and find the points of inflection (if any) for f(x) = x^3 - 3x + 2f(x)=x3−3x+2?

1 Answer
Nallasivam V
Oct 18, 2015

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

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

Explanation:

Given -

y=x^3-3x+2y=x3−3x+2

dy/dx=3x^2-3dydx=3x2−3

(d^2x)/(dx^2)=6xd2xdx2=6x

dy/dx=0 => 3x^2-3=0dydx=0⇒3x2−3=0

3x^2=33x2=3

x^2=3/3=1x2=33=1

sqrt(x^2)=+-sqrt1√x2=±√1

x=1x=1
x=-1x=−1

At >x=1x=1 ;

(d^2x)/(dx^2)=6(1)=6>0d2xdx2=6(1)=6>0

At >x=1 ; dy/dx=0;(d^2x)/(dx^2)>0 x=1;dydx=0;d2xdx2>0

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

At >x=-1x=−1 ;

(d^2x)/(dx^2)=6(-1)=-6<0d2xdx2=6(−1)=−6<0

At >x=-1 ; dy/dx=0;(d^2x)/(dx^2)<0 x=−1;dydx=0;d2xdx2<0

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

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

Watch this lesson also'on Maxima / Minima'

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