Caculus assignment

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Question 4, 6,7

It’s for my assignment and I need the full work.

Caculus assignment
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Borys_S
School: UIUC

The solutions are ready, please ask if something is unclear.docx and pdf files are identical

2. 5π‘₯ βˆ’ π‘₯ 3 𝑦 2 = 4𝑦 + 8, tangent line(s) at π‘₯ = 1.
For π‘₯ = 1 we obtain 5 βˆ’ 𝑦 2 = 4𝑦 + 8, 𝑦 2 + 4𝑦 + 3 = 0.
There are two solutions, 𝑦1 = βˆ’1, 𝑦2 = βˆ’3.
Differentiate with respect to π‘₯: 5 βˆ’ 3π‘₯ 2 𝑦 2 βˆ’ 2π‘₯ 3 𝑦𝑦 β€² = 4𝑦 β€² .
For π‘₯ = 1, 𝑦 = 𝑦1 = βˆ’1 we obtain 5 βˆ’ 3 + 2𝑦 β€² = 4𝑦 β€² , i.e. 𝑦 β€² = 1.
For π‘₯ = 1, 𝑦 = 𝑦2 = βˆ’3 we obtain 5 βˆ’ 27 + 6𝑦′ = 4𝑦 β€² , i.e. 𝑦 β€² = 11.
This way we have two points with known coordinates and tangent slopes.
The first line is 𝑦 = 1 βˆ™ (π‘₯ βˆ’ 1) βˆ’ 1 = 𝒙 βˆ’ 𝟐, the second is 𝑦 = 11 βˆ™ (π‘₯ βˆ’ 1) βˆ’ 3 = πŸπŸπ’™ βˆ’ πŸπŸ’.

4. 𝑦 = (2 βˆ’ 5π‘₯ 2 )4 (3π‘₯ 2 βˆ’ 8)5 ...

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