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I solved the problem twice. In the upper one, which is correct, I have the extreme points +3 and -3, and in the other one I have two times -3.
Why is that so I don't get the same result
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From the function F(p)=500-4p Is it 125?
I need to read the coefficient of quadratic equations here. why do I have to ignore the minus?
What is this function? So, in the individual quadrants?
is something different than
The former can also be written as (x+3)(x-3) (3rd binary formula)
The second is (x+3)2
Edit:
I don’t know what you were thinking. The function looks like this:
An extreme point at H(3|216). N(5|0) and S(-3|0) are the two zero points, the latter also being the saddle point (and thus the turning point). Further turning point at W(1|128)
Edit2:
Now I’ll see what you did.
You took the first derivation:
f'(x) = -2 x3 – 6 x2 + 18 x + 54 and 0 set.
and divided by -2:
x3 + 3 x2 – 9 x -27 = 0
This is (x+3)(x+3)(x-3) a double zero point at -3 and a simple one at +3
If you recreate the task twice, i.e. once by (x+3) and once by (x-3), you will receive the first time (x2-9) = (x+3)(x-3) and the second time (x2 + 6x + 9) = (x+3)2
In both cases f’ has a double zero at -3 and a simple at +3. You get no other result, but only in a different order.
You mean (x+3)(x-3)?
Sure. Correct, thank you.
In the second task, one does not see whether a minus or + is before 6x, so one cannot say the correct solution but if it is -6x, then the solution x1/2=3 and it is a double zero; at +6x is the solution x1/2=-3 and it is again a double zero;
you do not have an error in your 2nd invoice wthe discharge is correct;
But since you have calculated the same task, the derivation must be wrong;
f'(x)=-2x^3-6x^2-18x+54
you have completely different derivatives that do not fit the task
I checked it and my instructions are correct
I don’t understand your problem. You come to different results because it appears to be different functions.
In both invoices you could have dispensed with the pq formula, because you can see the solutions when looking at the functions. It’s simple binomes. Their solutions are obtained from the set of the zero product.
And at the second, it’s just a solution. Because a+0 and a-0 IMMER a results. This is because 0 is zero – the mathematician calls the 0 the neutral element of the addition.
What is the solution?
set of zero product.
In the first (x-3)*(x+3)=0 then you can see it.
In the second (x+3)*(x+3)=0, this is also trivial.
If you have a beautiful hammer (here PQ formula), everything looks like a nail. PQ formula always works. “Seeing solutions” requires thought and experience.
The set of zero product is also a beautiful and precise tool…. and much easier to understand and notice than the pq formula.
You have another function below? It’s clear that something else comes out or I don’t understand your problem?
at the top you have a zero at *3 and a zero at -3
at the lower one double zero point at -3 (that is, the graph only touches the x axis but does not pass)
Yes I have another function because I did Horner scheme with +3 instead —3. And it’s still right because at the end 0 comes, but actually I need the points + and – 3
Could you pack this text in sentences that you can understand?
Could you please do the whole job in the question?
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