For the function f(x) = g(x)=7x^3(9x^2-1)

Question

What are the critical points?

What are the local minima and maxima?

What are the inflection points?

Answer 1

g(x) = 7x^3(9x^2 - 1) = 63x^5 - 7x^3

There are 0s at 0 and +- 1/3. 0 is a triple root. Thus, we know that there is an inflection and there will be 2 additional in (-1/3, 1/3) before doing any calculations.

Now, let's find our critical points and points of inflection.

g'(x) = 315x^4 - 21x^2.

Factor to find zeroes.

g'(x) = 21x^2(15x^2 - 1)

Our critical points are 0 and +- √1/15 = 0, +-√15/15

We can tell from our understanding of the function that - √1/15 = - √15/15 is a maximum and √1/15 = √15/15 is a minimum and 0, which is a triple root of the original polynomial, is neither (consider the zeroes and their multiplicity and the behavior of the function for large and small x), or we can look at the second derivative there.

g''(x) = 1260x^3 - 42x = 42x(30x^2 - 1)

Thus, there will be an inflection at 0 and +- √1/30 or 0 and +- √30/30

We can calculate g'' at +- √1/15 if we like.

42x(30x^2 - 1) at - √1/15 is -42/√15 (30*( - √1/15)^2 - 1) = -42/√15 (30/15-1) = -42/√15 (1) = -42/√15 < 0, so we have a maximum.

42x(30x^2 - 1) at √1/15 = 42/√15 > 0, so we have a minimum.