Edexcel Core 2 calculus question. NEED HELP!!! Watch

StarGirlFrankie
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I've been given some core 2 calculus questions for homework for tomorrow, but I'm stuck on one!!! Here goes:

The figure above shows part of the curve C with equation; y=3x^1/2 - x^3/2
The point A on C is a stationary point and C cuts the x-axis at the point B.

a) B = (3,0)

b) Find the coordinates of A (the maximum point).

I've honestly tried everything I can think of! I tried to differentiate, but got 3/4(x^-1/2 - x^1/2)and couldn't find x from that.

Please help me!!!
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TeeEm
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where is everybody these days ...
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gdunne42
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(Original post by StarGirlFrankie)
I've been given some core 2 calculus questions for homework for tomorrow, but I'm stuck on one!!! Here goes:

The figure above shows part of the curve C with equation; y=3x^1/2 - x^3/2
The point A on C is a stationary point and C cuts the x-axis at the point B.

a) B = (3,0)

b) Find the coordinates of A (the maximum point).

I've honestly tried everything I can think of! I tried to differentiate, but got 3/4(x^-1/2 - x^1/2)and couldn't find x from that.

Please help me!!!
x^-1/2 - x^1/2 = 0

multiply through by x^1/2 to get rid of the fraction

or rearrange

1/root(x)=root(x)
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math42
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I don't see where you're getting 3/4 from, check that again (although it won't actually affect the answer)

You know that at the stationary point, dy/dx = 0. x^(-1/2) - x^(1/2) = 0. You can multiply by a certain power of x in order to turn this into a linear equation.
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Hinicetomeetyou
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(Original post by gdunne42)
x^-1/2 - x^1/2 = 0

multiply through by x^1/2 to get rid of the fraction

or rearrange

1/root(x)=root(x)
You've lost a 3 that was in the OP amongst this. I'm not sure we're supposed to give full solutions.

@OP, Differentiate again, your differential looks incorrect. Equate this to zero and solve for X. It is definitely possible.
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gdunne42
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(Original post by Hinicetomeetyou)
You've lost a 3 that was in the OP amongst this. I'm not sure we're supposed to give full solutions.
.
No I haven't, I assumed the OP meant to type 3/2(x^-1/2 - x^1/2) as it seems unlikely they would write their dy/dx in a factorised form otherwise
If 3/2(x^-1/2 - x^1/2) = 0
(x^-1/2 - x^1/2) = 0

I don't believe I have offered a full solution though I could have hinted that they should multiply through by some power of x to eliminate the fraction
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StarGirlFrankie
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(Original post by 13 1 20 8 42)
I don't see where you're getting 3/4 from, check that again (although it won't actually affect the answer)

You know that at the stationary point, dy/dx = 0. x^(-1/2) - x^(1/2) = 0. You can multiply by a certain power of x in order to turn this into a linear equation.
I've just realised that I typed it wrong, and meant 3/2, not 3/4. and do you mean times the power by something to make it into a linear equation, or make it 1/sqrtx - sqrtx?
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StarGirlFrankie
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(Original post by StarGirlFrankie)
I've just realised that I typed it wrong, and meant 3/2, not 3/4. and do you mean times the power by something to make it into a linear equation, or make it 1/sqrtx - sqrtx?
oh wait... i think I just realised... woops!
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Hinicetomeetyou
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(Original post by gdunne42)
No I haven't, I assumed the OP meant to type 3/2(x^-1/2 - x^1/2) as it seems unlikely they would write their dy/dx in a factorised form otherwise
If 3/2(x^-1/2 - x^1/2) = 0
(x^-1/2 - x^1/2) = 0

I don't believe I have offered a full solution though I could have hinted that they should multiply through by some power of x to eliminate the fraction
I'm talking about in the original equation, going from that to your differential.
y=3x^1/2 - x^3/2

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gdunne42
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(Original post by Hinicetomeetyou)
I'm talking about in the original equation, going from that to your differential.
y=3x^1/2 - x^3/2

dy/dx= 3/2x^-1/2 - 3/2x^1/2 = 3/2(x^-1/2 - x^1/2)


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