Maths question

Hi, please could I have help on this question?
To prove a point of inflection, I’ve searched online and it says you plug in values around the point of infleciton into d2y/dx2 and see if there is a sign change. For question 14c of this paper , the ms does not do that so I’m really confused now? Please could I have some clarification on this?
Paper: https://pmt.physicsandmathstutor.com/download/Maths/A-level/Papers/OCR-MEI/Paper-2/QP/Nov%202021%20QP.pdf
Ms: https://pmt.physicsandmathstutor.com/download/Maths/A-level/Papers/OCR-MEI/Paper-2/MS/Nov%202021%20MS.pdf
Thank you!

14b/c) asks about the stationary points, not the points of inflection.
The question gives you the 2nd dervative in factorized form so its easy to see its zero/inconclusive at the stationary poinit at x=2.
The ms does two similar things
1) it shows the gradient is positive at x=1 and 3, so the sign of hte gradient doesnt change.
2) it shows the curve is increasing between x=1 and 2, and between x=2 and 3 which is the same as showing the gradient is positve/doesnt change in those two intervals.
Either is fine.

but if d2y/dx2 is 0, does that not mean it could be an inflection point so is that not what they have tested for using the two ways? The first way they have done it, you have said that the gradient does not change. So is that how we prove a point of inflection or is it only by subbing into d2y/dx2? thanks!

The question asks about stationary points, not points of inflection. Like f(x)=(x-2)^3, the given function has a stationary point at x=2 (f'(2)=0) which is also as a point of inflection (f''(2)=0) but calling it a point of inflection is irrelevant. The only important thing is you cant use the usual second derivative test to determine min or max or .... So you look at the gradient sign (change or not) either side of the stationary point, or equivalently compare the function values.

ohh i get what you are saying, but i'm unsure why we can't use the second derivative test?

Because the 2nd derivative is zero. So it could be a min or max or a point of inflection or ... In this case, the second derivative changes sign at x=2 and its a point of inflection / function is increasing / ... The ms does this by showing the function is increasing at that point, but its fairly simple to argue that f'' changes sign if you really wanted to as f'' is given to you in factorised form.

sorry i meant why can't we use values near to x=2 and input into the second derivative to show that there is a change of sign?

You could do. Just like you could argue the factorised form is (roughly)
f''(x) = (x-2)g(x)
which must change sign at x=2 because (x-2) changes sign but g(x) doesnt. The guidance (last column) says that considering the sign of f'' either side of x=2 is ok with a suitable explanation.

oh ok i didn't see that, thank you!

Just be clear about the with suitable explanation part. Usually the simplest (least calculus) explanation/approach is best to make sure you get the marks.