Another maths question!

To find the points in the first place do dy/dx of the curve = to 0 and then that’s ur x coords then plug that in to work out y

Thx i understand that but dont know how to do dy/dx for b?

Thx i understand that but dont know how to do dy/dx for b?

Thx i understand that but dont know how to do dy/dx for b?

U work out the dy/dx of a point on each side and then work out whether it’s a u curve or n curve from that

Differentiate and set y' = 0 to get A: y' = 4x^3 - 16x = 4x(x+2)(x-2), giving x = 0, 2, -2
Differentiate again to find y''= 12x^2 - 16

x = 0 gives y'' = -16, < 0 so maximum
x = 2 gives y'' = 32, > 0 so minimum
x= -2 gives y'' = 32, > 0 so minimum

The curve is a w shape.
For the record, setting y'' = 0 gives the x-coordinates of the points of inflection.

As for B, . it's unclear whether you mean x - 6^0.5 or (x-6)^0.5

y = x - 6^0.5 is a straight line with y'=1, so no stationary points anyway
y = (x-6)^0.5 with y'= 0.5(x-6)^-0.5 which is always positive and never = 0, so it's is an increasing function (draw it on desmos), so also no stationary points.

Not if the question specifically says use the second derivative. You'd get 0 marks if you did it your way I'm afraid.

It doesn’t say use the 2nd derivative tho... or r u on about a specific Q from an exam?? If so u did a different one to me

... what is a good palce for revision and understandable< i literally failed

Er, "By evaluating d^2y/dx^2 ( d squared Y over DX squared ) at each stationary point, determine whether it is a minimum or maximum point."

U can identify by taking the dy/dx of each side and if then by working with that u can tell whether the curve was shaped like an n or a u

Indeed you can, and this is particularly useful to determine whether a function is increasing or decreasing around a point of inflection. It's also the only realistic way of doing it if differentiating y' would be somewhat onerous!!!

However, if an exam question told you specifically to use the 2nd derivative (as the questions in the op do), then you'd have to do so if you wanted any marks.

Don't shoot the messenger

Differentiate and set y' = 0 to get A: y' = 4x^3 - 16x = 4x(x+2)(x-2), giving x = 0, 2, -2
Differentiate again to find y''= 12x^2 - 16

x = 0 gives y'' = -16, < 0 so maximum
x = 2 gives y'' = 32, > 0 so minimum
x= -2 gives y'' = 32, > 0 so minimum

The curve is a w shape.
For the record, setting y'' = 0 gives the x-coordinates of the points of inflection.

As for B, . it's unclear whether you mean x - 6^0.5 or (x-6)^0.5

y = x - 6^0.5 is a straight line with y'=1, so no stationary points anyway
y = (x-6)^0.5 with y'= 0.5(x-6)^-0.5 which is always positive and never = 0, so it's is an increasing function (draw it on desmos), so also no stationary points.

Ok np I just do whatever I see in the question the easiest way possible lol maybe I didn’t quite understand it properly due to it being payed out on a screen style. and I’m only gcse so... I’d probably get away with it 😝

Fair enough. I'm a maths teacher so have a bit of an advantage