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i am stuck on this question
Find the x co-ordinates of the points on the curve y= (x+1)(x-2)^2 at which the gradient is zero. test whether the points are local maxima or minima.
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Hi Joker;
Differentiate your function to get:
Now set that to 0 and solve:
Solve by factoring: You can pull out a factor of (x-2) from both terms.
x - 2 = 0 so x = 2
(x-2) + 2(x+1) = 0
(x-2) + 2x + 2 = 0
3x = 0 so x = 0 You have 2 roots x = 2 and x = 0.
Can you finish from here?
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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Hi Joker;
Good, use this formula to differentiate the function:
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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Yes, at both local and global maxima and minima the slope is 0.
Notice I use the word slope as a opposed to gradient. Gradient is a vector while slope or grade is a scalar.
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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Were you able to do the differentiation, that is above?
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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Hi;
Did you follow that?
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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Hi Joker;
Don't worry, we will go through it step by step:
Hav you seen the above formula for diff. a product?
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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Okay, say
f(x) = (x+1)
g(x) = (x-2)^2
Now plug in to the formula and what do you get?
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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something like this?
d(x+1)(x-2)^2/dx = (x+1)2(x-2) + (x-2)^2
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Hi; That is correct.
To determine whether the points are maxima or minima you need the 2nd derivative. So differentiate your answer again, and let me see what you get.
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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Yes, or any other way, you could expand it if you want. They are both good ways.
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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Hi Joker;
That is not correct. Let me see what you did.
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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Hi Joker;
(x+1)2(x-2) + (x-2)^2 is only a second degree poly, when you differentiate it it will be smaller a first degree, not larger.
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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o thought so, how about this?
i expanded this first (x +1) (2x-4) = 2x^2 - 4x + 2x -4
then
(x - 2)(x-2) = x^2 + 4
so 2x^2 - 2x - 4 + x^2 + 4 = 3x^2 -2x
f'(X) = 6x -2
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Hi;
(x +1) (2x-4) what happened to the 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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(x+1)2(x-2) = (x+1)(2x-4)
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But it is x+2 not x + 1.
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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Your (x-2)(x-2) is wrong.
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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(x-2)(x-2)= x^2 - 4x + 4 not x^2 + 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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Hi;
Sorry, that was a mistake on my part, I thought that's where you went wrong, but it was on this:
(x-2)(x-2)= x^2 - 4x + 4 not x^2 + 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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ah ok
so its meant to be 6x-6?
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That's it!! Now you have the 2nd derivative. Do you know what you do with 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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If f''(x) = 0 then it tells you nothing about the points.
Do you remember what your 2 critical points are?
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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x = 0 and x = 2?
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Now plug them into the second derivative and you get
6x - 6 with x = 0 = - 6 so this is a local max.
6x - 6 with x = 2 = 6 so this is a local min.
Now check by graphing the original function.
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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