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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)

Nehushtan

Member

Registered: 2013-03-09

Posts: 957

Re: Calculus

Solve what?

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meerblau

Member

Registered: 2016-02-17

Posts: 3

Re: Calculus

Sorry, I edited the post.

I would like to know what z is.

zetafunc

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Registered: 2014-05-21

Posts: 2,436

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Re: Calculus

Did this question come with any other information?

Grantingriver

Member

Registered: 2016-02-01

Posts: 129

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.

bobbym

bumpkin

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.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
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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 ;-).