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

Question

Emily Ammerman · Accepted Answer

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]}

Watch this lesson also'on Maxima / Minima'

Isaiah Allen · Answer

undefined To describe the concavity of the graph of $ f(x) = x^3 - 3x + 2 $ and find the points of inflection:

1. Calculate the second derivative of $ f(x) $ to determine the concavity.
2. Find the points where the concavity changes, which are the points of inflection.

First derivative of $ f(x) $:
$$ f'(x) = 3x^2 - 3 $$

Second derivative of $ f(x) $:
$$ f''(x) = 6x $$

Set $ f''(x) = 0 $ to find possible points of inflection:
$$ 6x = 0 $$
$$ x = 0 $$

Now, test the sign of $ f''(x) $ in intervals around $ x = 0 $ to determine the concavity:

- For $ x < 0 $: Choose $ x = -1 $ (test point)
$$ f''(-1) = 6(-1) = -6 $$ (negative)
- For $ 0 < x $: Choose $ x = 1 $ (test point)
$$ f''(1) = 6(1) = 6 $$ (positive)

Since the concavity changes from concave down (negative) to concave up (positive) at $ x = 0 $, there is a point of inflection at $ x = 0 $.

Therefore, the graph of $ f(x) = x^3 - 3x + 2 $ is concave down for $ x < 0 $, concave up for $ x > 0 $, and has a point of inflection at $ x = 0 $.