How do you sketch the graph #y=x^4-2x^3+2x# using the first and second derivatives?

Question

Ezekiel Alexander · Accepted Answer

See process below

1.- Domain is #RR#
2.- Analyze the roots of equation #x^4-2x^3+2x=0# in order to find axis intercepts

Notice that #x^4-2x^3+2x=x(x^3-2x^2+2)=0# thus #(0,0)# is a passing point. There is no more integer roots

3.- First derivative #f´(x)=4x^3-6x^2+2=(x-1)^2(4x+2)#

Analyze the sign of derivative. #(x-1)^2# is allways positive, then the sign of derivative dependes only of sign of #4x+2#

#4x+2>=0#
#x>=-1/2#

Then function is increasing in #(-1/2,oo)# and decreasing in #(-oo,-1/2)# then in #x=-1/2# has a minimum

Derivative is zero in #x=-1/2# and #x=1# but #x=1# is not a maximum neither a minimum because

#f´´(x)=12x^2-12x=12x(x-1)#. then #f´´# become #0# only in #x=0# and #x=1#. Thus inflection points.

A sketch is presented below

Aurora Alvarado · Answer

undefined To sketch the graph of $ y = x^4 - 2x^3 + 2x $ using the first and second derivatives:

1. Find the first derivative of $ y $ with respect to $ x $ and locate the critical points.
2. Find the second derivative of $ y $ with respect to $ x $ and determine the concavity of the function.
3. Locate the inflection points by setting the second derivative equal to zero and solving for $ x $.
4. Plot the critical points, inflection points, and any intercepts of the function.
5. Sketch the graph, indicating increasing/decreasing intervals based on the first derivative and concave up/down intervals based on the second derivative.