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#1 2016-02-17 05:19:13

meerblau
Member
Registered: 2016-02-17
Posts: 3

Calculus

My math classes are toooo long ago. Any help to sove this? :-)


f(z) = 4z4-3z3+15z2-9
f(y) = 7y4+38y3+7y2-200
f’(y) = 29772
f’(z) = -1+873y

Last edited by meerblau (2016-02-17 05:38:36)

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#2 2016-02-17 05:30:00

Nehushtan
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Registered: 2013-03-09
Posts: 905
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Re: Calculus

Solve what?


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#3 2016-02-17 05:39:30

meerblau
Member
Registered: 2016-02-17
Posts: 3

Re: Calculus

Sorry, I edited the post.

I would like to know what z is.

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#4 2016-02-17 07:11:47

zetafunc
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Registered: 2014-05-21
Posts: 2,109
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Re: Calculus

Did this question come with any other information?

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#5 2016-02-17 14:21:52

Grantingriver
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Registered: 2016-02-01
Posts: 53

Re: Calculus

f(z) = 4z⁴-3z3³+15z²-9.  → 1
f(y) = 7y⁴+38y³+7y²-200 → 2
f’(y) = 29772  → 3
f’(z) = -1+873y → 4
From the first and the second equations we have:
f'(z)=16z3³-9z²+30z and f'(y)=28y³+114y2²+14y hence:
2(14y³+57y²+7y)=29772 ⇒ 14y³+57y²+7y= 14886 ⇒ y=9, y≈-6.5+8.7i and y≈-6.5-8.7i
If you seek the real solution then we use the first real answer which is "y=9" in the forth equation so we have:
16z³-9z²+30z=-1+873×9 ⇒ 16z³-9z²+30z=7856 ⇒ z=8, z≈6.9+3.7i and z≈-6.9+3.7i.
Now since it is obvious that you need the real roots (since they are the required values in many applications), therefore the solutions are "y=9 and z=8".

Note: you can find other solutions by substituting the complex roots of y in the forth equation.

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#6 2016-02-17 14:34:37

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Calculus

Hi;

That looks right, sorry for the confusion caused by my answer. I have deleted 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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#7 2016-02-17 20:11:38

meerblau
Member
Registered: 2016-02-17
Posts: 3

Re: Calculus

Thank you so much for your help! Now I wonder even more that I was able to sollte that during my school days ;-).

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