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

Answer 1

Samantha Adkison

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]}

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Answer 2

Charles Anctil

To sketch $y = x^4 - 2x$ using the first and second derivatives:

Find the first derivative: $y' = 4x^3 - 2$ .
Find critical points by setting $y' = 0$ and solving for $x$ .
Use the second derivative test to determine the nature of critical points.
Determine concavity intervals using the second derivative.
Sketch the graph based on the information obtained from steps 2-4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.