How do you use the first and second derivatives to sketch #y=x^4-2x#?

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Answer 1 · 2017-03-30T13:18:43

minimum #(0.79;-1.19)#
inflection #(0;0)#
convex

The first derivative is:

#f'(x)=4x^3-2#

Let's study the solution of the inequality

#f'(x)>=0#

that's

#4x^3-2>=0#

#x>=root(3)(1/2)~=0.79#

It means that

if #x < root(3)(1/2)# the function is decreasing;

if #x > root(3)(1/2)# the function is increasing;

then if x=if #x < root(3)(1/2)# the function has a minimum and there #f(x)=(root(3)(1/2))^4-2*root(3)(1/2)=(root(3)(1/16))-2*root(3)(1/2)~=-1.19#;

The second derivative is:

#f''(x)=12x^2#

Let's study the solution of the inequality

#f''(x)>=0#

that's

#12x^2>=0# that is verified #AAx in RR#

It tells the function is convex and if #x=0# there is a point of inflection and it is the origin #O(0;0)#

graph{y=x^4-2x [-2, 3, -2, 5]}