# Carefully skecth the graph of f(x)= x^4-6x^2, clearly showing and labeling the x & y coordinates of all absolute extrema, all local extrema... Carefully skecth the graph of f(x)= x^4-6x^2, clearly showing and labeling the x & y coordinates of all absolute extrema, all local extrema, and all inflection points . Justify your answer using calculus.

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Given `f(x)=x^4-6x^2` :

`f'(x)=4x^3-12x`

`f''(x)=12x^2-12`

(a) All extrema occur at critical points (i.e. where f'(x)=0 or f'(x) fails to exist.) Since this is a polynomial, we are only concerned with finding points where f'(x)=0:

`4x^3-12x=0==>4x(x^2-3)=0` so x=0 or `x=+-sqrt(3)`

The graph of a quartic with positive leading coefficient has end behavior of `lim_(x->+-oo)=oo` so there will be no absolute maximum. There will be an absolute minimum.

We test values of f'(x) on the following intervals:

`(-oo,-sqrt(3):` f'(x)<0 so the function is decreasing on this interval.

`f(-sqrt(3))=-9`

`(-sqrt(3),0):` f'(x)>0 so the function is increasing on this interval.

f(0)=0

`(0,sqrt(3)):` f'(x)<0 so the function is decreasing on this interval

`f(sqrt(3))=-9`

`(sqrt(3),oo):` f'(x)>0 so the function is increasing on this interval.

There is a local maximum at x=0 with value 0, and the absolute minimum for the function occurs at `x=-sqrt(3),x=sqrt(3)` with value -9.

(b) Points of inflection occur when the concavity of the graph changes at a point -- i.e. the sign of the secon derivative changes at the point.

`12x^2-12=0==>x=+-1` . We check the sign of the second derivative on the following intervals:

`(-oo,-1):` f''(x)>0 so the function is concave up on this interval.

`(-1,1):` f''(x)<0 so the function is concave down on this interval

`(1,oo):` f''(x)>0 so the function is concave up on this interval.

The points of inflection are at x=-1 and x=1.

The y-intercept is at 0; the x-intercepts are `x=0,+-sqrt(6)`

The graph:

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