Yep

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.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.

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

Spoiler:

A point of inflection is when and a maximum is when

A minimum is when [tex]\dfrac{d^2y}{dx^2} > 0.Then next part asks:

(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

I can see no other way to plug x=a and solve for appropriate values of a. Perhaps someone else will know better?