# Quick DifferentialWatch

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#1
From a C4 book, integration chapter but this is C2-level differentiation.

Show that the curve has a minimum point and find it's coordinates.

Book says its (1/2,3) but I keep getting x=8 and x=0. I'm pretty sure I'm not going wrong, but it's been a while since a differential this simple .

Can someone point out my silly mistakes??

Thanks.
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11 years ago
#2
Is it not a half?

dy/dx = (-x^-2) + 8x = 0
Add the negative bit to the other side, multiply both sides by x^2 and rearrange for x.
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#3
A half is what it is in the back of the book, I keep getting 8 :/
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11 years ago
#4
(Original post by samson89)
A half is what it is in the back of the book, I keep getting 8 :/
You should get to the point in my last post where x^3 = 1/8 which implies x = 1/2?
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11 years ago
#5
Write out your working so we can see where you went wrong.
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11 years ago
#6
(Original post by notnek_01)
Write out your working so we can see where you went wrong.
I greatly enjoyed your signature spoiler. Thanks.
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#7
Thanks Colin, uneleivable the mistakes I can make sometimes...

An A in As level maths and I cant see that x^-2 is 1/x^2 and not -x^2.

Can we pretend this thread never existed?
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11 years ago
#8
Dude you had some time on your hands to do those spoilers lol
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11 years ago
#9
(Original post by ColinOfEdinburgh)
I greatly enjoyed your signature spoiler. Thanks.
You're welcome.
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11 years ago
#10
(Original post by samson89)
Can we pretend this thread never existed?
I'm an honours year electrical engineer and still forget Ohm's law.
So yes, I'm sure we can forget this all!
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