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Okay, do you see what the curves look like? Do you understand it? Do you see why they are max and min?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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ops i mean min i get x=2 y=8.77 x 10^-24
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That is round off error. The true answer is 0 which you can prove by plugging 0 into your original equation.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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yep i understand why they a max and min but i dont get what to do with the -6 and 6
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Look at post # 35 for the rules.
When you plug in the critical points into the 2nd der and it is less than 0 than that is a local maximum,
When you plug in the critical points into the 2nd der and it is greter than 0 than that is a local minimum,
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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so x=2 is min and x=0 is max?
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From the graph you can see that and you also shown that by using calculus. How was it?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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sweet thx so much
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Glad to help, bring more problems.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hmmm i forgot theres also a part b to the question find the x co-ordinate of the point at which the tangent is parallel to the tangent at the point.
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darn forgot the points (3,4)
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Hi Joker;
That point is on the curve y = (x+1)(x-2)^2 do you want the tangent to it?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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umm the whole question goes like this,
A) find the x co-ordinate of the points on the curve Y = (x+1)(x-2)^2 at which the gradient is zero. test whether the points are minima or maxima. this part with your help is done. thx agian
B) Also find the x co-ordinate of the point at which tangent is parallel to the tangent at the point (3,4)
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Hi Joker;
The tangent at (3,4) on the curve Y = (x+1)(x-2)^2 has the equation y = 9x - 23, I do not understand do you want another tangent at (0,4)?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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sry just thinking about it, i dont really get the question either the "tangent is parallel to the tangent at the point (3,4)" confuses me dont worry about i will get my lecturer the explain thx for the help.
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Hi;
I got it. They want the tangent that is parallel to that one. Do you know how I got y= 9x -23 ?
Start at the point (3,4) and plug that into the first derivative.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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plug (3,4) into f'(x) = (x +1)2(x+1) + (x-2)^2?
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Yes, that will give you the slope of the original function at the point (3,4).
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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umm where doe the 4 go?
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Hi;
On the left, remember it is:
y = (x -2)2(x+1) + (x-2)^2 because y = f(x),
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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29?
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Hi;
You are working with the wrong equation: Plug into this to get 9:
(x -2)2(x+1) + (x-2)^2
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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where did that equation come from? wasnt this the first derivative f'(X) = (x + 1)2(x+1) + (x-2)^2
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Nope, check post # 2
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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4 = (3-2)^2 + 2(3+1)(3-2)
4 = (3-2)(3-2) + (6+2)(3-2)
9-6-6+4+24-12+6-4-4= 11
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