C3 minimum and maximum gradients

A key point to remember is dy/dx represents the gradient of the tangent (or slope) of the original curve and that d2y/dx2 represents the rate of change of the gradient of the tangent.

Therefore, similarly to finding the maximum or minimum values of y on a graph by setting dy/dx = 0, to find the minimum or maximum gradient you set d2y/dx2 = 0 (if the second derivative gave two values for x, you would check d3y/dx3 to check if they're a minimum or maximum, i.e. d3y/dx3 > 0, gradient is a minimum value, d3y/dx3 < 0, gradient is a maximum value).

I've put my answer to the tangent below, but don't check it until you've done it fully yourself so you can have a go first.

Woo, I got it right! Really good explanation, cheers!

Can you help me with Q9 as well please lol?
9iv) verify by calculation that the value of x at the point of intersection of y= tan^-1(x) and y= 2tan^-1 (x-1) is 1.54

The Markscheme seemed to use a sort of "change of sign" method plugging in 1.535 and 1.545, although there was no change of sign. i tried equating them but that didn't seem to work

No worries!

For that question, all you have to do is equate both (as you did), and you should get 2tan^-1(x-1) - tan^-1(x) = 0.

Now since the function equals zero, you plug in those two values [1.535, 1.545], because you want to prove that the root is 1.54 to 3 s.f., and each value should yield a corresponding (small) value with a different sign.

Subbing in 1.535 you get quite a small negative value, whilst with 1.545 you get quite a small positive value, so that proves that the root of the intersection is 1.54 to 3 significant figures

Alternatively, you could do tan^-1(x) - 2tan^-1(x-1) = 0, where subbing in 1.535 will yield a small positive value and 1.545 will yield a small negative value, that would be sufficient.

Oh wow that was annoyingly obvious lol thanks. I'm dreading this exam, are you doing it as well?

I do the Edexcel syllabus

I prefer the OCR syllabus though, mainly because papers are harder and therefore grade boundaries are generally fairer. Edexcel papers are easy, problem is that a minor error can push you down quite a few UMS sadly.

If you ever need help just ask on the forum, there are loads of people here who are willing to help

Fair point, I've done a few C2 and C3 papers from Edexcel and they were easier. Cheers mate, how do you go about proving trig identites? I waste so much time trying all 4 of the methods (left to right, right to left, division to get 1, and subtracting to get 0) until I find the answer

What do you mean, proving trig identities such as 1 + tan^2(x) = sec^2(x) and 1 + cot^2(x) = cosec^2(x)?

My bad, for example: Prove that CotA-TanA= 2cot2A
Is there a general "rule" to follow as in I proved the identity above but on my 3rd attempt :/

Always prove from a side where you think it'll be most easiest to prove from. For this one, I'd prove from the RHS.

For instance, the RHS is the same as 2cos2A/sin2A, which is the same as 2(cos^2(A) - sin^2(A))/2sinAcosA, which you can split up as being cos^2(A)/sinAcosA - sin^2A/sinAcosA, which yields cotA - tanA on the LHS.

You eventually start seeing what you have to do before you've done it, just takes a little practice.