Maths Core 2 AS level anyone good with calculus to find set of values of x for f(x)?

2)

y = 4x^2 + 1/x

y = 4x^2 + x^-1

dy/dx = 4(2)x^1 + (-1)x^(-1-1)

dy/dx = 8x -1/x^2

At the stationary point the gradient is '0' (zero)

So dy/dx = 0

Hence

8x - x^-2 = 0

8x = x^-2

8x^3 = 1

x^3 = 1/8

x = (1/8)^(1/3)

x = 1/2

the superb thank you to do it somewhat is to "complete the sq.." as quickly as you are trying this, it is going to likely be waiting to coach you the minimum component and the x value for which this happens. in the beginning, enable f(x) = x^2-4x-12. Then, to end the sq., you ought to halve the integer next to the x. So, subsequently, -4 is going to -2. then you definately ought to placed this right into a squared bracket with an x- so we get (x-2)^2. whilst we improve this, we get x^2-4x+4. the priority is we've a +4 we don't choose! The question for sure shows that your 'C' value is -12. for this reason, do away with this +4 by minusing it. you ought to usual get (x-2)^2 -4 -12 it somewhat is (x-2)^2 - sixteen. From this, all of us understand the minimum component is at (2, -sixteen.) word- no longer -2. you ought to opposite the sign interior the bracket to get your x co-ordinate. A shop on with up question in a C1 examination ought to be to describe the modifications of the unique x^2 graph- this technique is nice!