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]}

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