c1 differentiation tangents Watch

mrdetermination
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#1
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it asks, find the coordinates of the point on the following curves where the gradient has the started value

y=3x^-6x+4 gradient is 0


this is really baffling me i keep getting the answer wrong can some one show me how it is done?
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gdunne42
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What have you done?
Show your working and the result you achieved and it will be easy to help you find where you are going wrong

It's about gradients of a curve so start by differentiation, then how can you use the result to find the value of x where the gradient = 0

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TenOfThem
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(Original post by mrdetermination)
it asks, find the coordinates of the point on the following curves where the gradient has the started value

y=3x^-6x+4 gradient is 0


this is really baffling me i keep getting the answer wrong can some one show me how it is done?
I assume that is 3x^2

Looks very straightforward

Can you show your working - if you are not able to solve this you must be making a basic error or have a basic mis-understanding - I will look at your working to see what the issue is
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Ranibizumab
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First, differentiate.

Then set dy/dx as zero

Solve for x

The value(s) or x where the gradient is zero is the x co-ordinate of the point you are looking for

Using this value of x, substitute it into the original equation to get the y value at this point.
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mrdetermination
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y=3x^2 - 6x+4 gradiant 0
dy/dx= 6x - 6=0

6x=6
x=6/6
x=1

y=6x1-6
y=0


it says the answer is 1,1
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Ranibizumab
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(Original post by mrdetermination)
y=3x^2 - 6x+4 gradiant 0
dy/dx= 6x - 6=0

6x=6
x=6/6
x=1

y=6x1-6
y=0


it says the answer is 1,1
Your method is correct, but you are using the x value of 1 in the wrong equation.
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THEMathlete
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(Original post by mrdetermination)
y=3x^2 - 6x+4 gradiant 0
dy/dx= 6x - 6=0

6x=6
x=6/6
x=1

y=6x1-6
y=0


it says the answer is 1,1
You need to plug x = 1 into the original equation...

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mrdetermination
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ow right i understand now thank you guys
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