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# Calculus/Help

Question
I very much like your "Instructions from the Expert"; had a chuckle at that one. I am homeschooling my son in Calculus, but I am having a little trouble with the lesson material; Particularly this type of problem; am hoping some good examples help.

Anyways, my question is:

Given f(x)=x^4-72x^2-17
Identify the inflection points and the intervals where the function is concave up or down.
Thanks,
John

Given the funciton is f(x) = x^4 - 72x^2 - 17, it can be seen that
f'(x) = 4x^3 - 144x and f"(x) = 12x^2 - 144.

Setting the 1st derivative to 0 would give 4x^3 - 144x = 0, or 4x(x^2 - 36) = 0.
The zeros of this are x=0, -6, and 6.

Setting the second derivative to 0 would give 12x^2 = 144, so x^2 = 12, so x = +/- root(12).

Using this, it means the iflection points (where f"=0) roots are at +/- root(12).
It means the critical points (f'=0) are at -6, 0, and 6.

Since the sign of the 2nd derivative tells whether the function is concave up or concave down,
we need to look at the intervals (-infinity,-root(12)), (-root(12),(root(12)), and
(root(12),infinity).

Since the second dervative is f"(x) = 12x^2 - 144, at the 1st and last intervals, this can be seen to be positive and in the middle interval it is negative.  This means the function is concave up at both ends and concave down in the middle.

Calculus

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