The Student Room Group
Reply 1
What have you got for the derivative of the function? Your error is likely to be there. If you post your workings I'll tell you where you have gone wrong
Reply 2
Original post by JSG29
What have you got for the derivative of the function? Your error is likely to be there. If you post your workings I'll tell you where you have gone wrong


8x - x^-2
Reply 3
Original post by esmeralda123
8x - x^-2


That's correct, what was your next step?
Original post by esmeralda123
8x - x^-2


So set it to zero and solve for xx.

Show your steps, since that's where you went wrong.
Dy/dx= 8x- 1/x^2= x(8-1/x^3)
X=0, or 8-1/x^3=0
8=1/x^3
1/8=x^3
X=1/2
Reply 6
Original post by Lechatn0ir
Dy/dx= 8x- 1/x^2= x(8-1/x^3)
X=0, or 8-1/x^3=0
8=1/x^3
1/8=x^3
X=1/2

Why does it become 1/x^3 when you take a factor of x out
dy/dx = 8x - x^-2

Set to = 0

8x^3 = 1
1/8 = x^3
so 1/2 = x
Original post by esmeralda123
Why does it become 1/x^3 when you take a factor of x out


There's no factor of x out...

1x3\frac{1}{x^3} = 88

What would x be?

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