Question

Find the derivative at each critical point and determine the local extreme values of y=x^2/3(x^2-4); x≥0.

Answer 1

Okay, you should get working on that derivative. What's your plan?

Answer 2

Not a clue honestly. I tried to derive it using the product rule and it's safe to safe that I failed at that. So now I figured to distribute the x^2/3 so working on that now!

Answer 3

You're saying the Quotient Rule didn't cross your mind? It is a division problem.

Answer 4

Wait, is it \(\dfrac{x^{2}}{3}(x^{2} - 4)\) or maybe \(\dfrac{x^{2}}{3(x^{2} - 4)}\)?

Answer 5

It may help just to get the 3 out of the way.

\(\dfrac{d}{dx}\dfrac{x^{2}}{3}(x^{2} - 4) = \dfrac{1}{3}\dfrac{d}{dx} x^{2}(x^{2} - 4)\)

You can work on the parts with variables and just keep the 1/3 along for the ride.

Answer 6

I see it as a multiplication problem. I'm not quite sure how you got the 3 in the denominator.

Answer 7

Isn't that what "/3" means?

Answer 8

Yes that is true, I don't know how to write it out all fancy like you. /3 is part of 2/3. x raised to the two thirds. It's a fractional exponent that just belongs to the x.

Answer 9

(x^2/3)(x^2-2) maybe makes more sense?

Answer 10

(x^2/3)(x^2-4) is actually the correct equation.

Answer 11

Please remember your Order of Operations.

x^2/3 = \(x^{2}/3 = \dfrac{x^{2}}{3}\)

x^(2/3) = \(x^{2/3}\)

Multiplying through should work just fine.

[x^(2/3)]*(x^2 - 2) = x^(8/3) - 2x^(2/3)

Okay, now what?