# Math Help - chain rule and stationary points question

1. ## chain rule and stationary points question

I am a bit stuck on a chain rule question and would like to ask for some advice.

The question starts:

Q. The graph of y = (x^3 - x^2 +2)^3, is shown (point of inflection at x = -1, max at x=0 and min at unknown x=a point.

OK so I use chain rule and get dy/dx = 3(x^3 - x^2 +2)^2 (3x^2 - 2x)

Then next part of question is:

(ii) Verify, showing your working clearly, that when x = -1, the curve has a point of inflection and when x = 0, the curve has a maximum.

I tackled this by saying stationary point when dy/dx = 0, then calculated that when x = -1, dy/dx does = 0. Same for x = 0.

But would I need to do 2nd order differential - ie d^2y/dx^2? Is that required. On above dy/dx is quite hard. Just want to know if there is an easier way.

(iii) the curve has a minimum when x = a. Find a and verify that this corresponds to a minimum.

solving this for dy/dx = 0 is quite tricky. How do I solve that? Or is there a way I am missing?

Angus

2. ## Re: chain rule and stationary points question

Originally Posted by angypangy
I am a bit stuck on a chain rule question and would like to ask for some advice.

The question starts:

Q. The graph of y = (x^3 - x^2 +2)^3, is shown (point of inflection at x = -1, max at x=0 and min at unknown x=a point.

OK so I use chain rule and get dy/dx = 3(x^3 - x^2 +2)^2 (3x^2 - 2x)
Yep

Then next part of question is:

(ii) Verify, showing your working clearly, that when x = -1, the curve has a point of inflection and when x = 0, the curve has a maximum.

I tackled this by saying stationary point when dy/dx = 0, then calculated that when x = -1, dy/dx does = 0. Same for x = 0.

But would I need to do 2nd order differential - ie d^2y/dx^2? Is that required. On above dy/dx is quite hard. Just want to know if there is an easier way.
You'd need to take the second derivative. Thus far you've only proved that -1 is a stationary point, not a point of inflection.

To take the second derivative use the product rule and the chain rule

Spoiler:
$u = 3(x^3-x^2+2)^2 \rightarrow u' = 6(x^3-x^2+2)(3x^2-2x)$

$v = 3x^2-2x \rightarrow v' = 6x - 2 = 2(3x-1)$

$\dfrac{d^2y}{dx^2} = 6(3x^2-2x)^2(x^3-x^2+2) + 6(3x-1)(x^3-x^2+2)^2$

You could simplify but not much point since you're only going to plug and chug

A point of inflection is when $\dfrac{d^2y}{dx^2} = 0$ and a maximum is when $\dfrac{d^2y}{dx^2} < 0$