 # Caculus assignment Anonymous

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Question 4, 6,7

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Borys_S
School: UIUC  2. 5𝑥 − 𝑥 3 𝑦 2 = 4𝑦 + 8, tangent line(s) at 𝑥 = 1.
For 𝑥 = 1 we obtain 5 − 𝑦 2 = 4𝑦 + 8, 𝑦 2 + 4𝑦 + 3 = 0.
There are two solutions, 𝑦1 = −1, 𝑦2 = −3.
Differentiate with respect to 𝑥: 5 − 3𝑥 2 𝑦 2 − 2𝑥 3 𝑦𝑦 ′ = 4𝑦 ′ .
For 𝑥 = 1, 𝑦 = 𝑦1 = −1 we obtain 5 − 3 + 2𝑦 ′ = 4𝑦 ′ , i.e. 𝑦 ′ = 1.
For 𝑥 = 1, 𝑦 = 𝑦2 = −3 we obtain 5 − 27 + 6𝑦′ = 4𝑦 ′ , i.e. 𝑦 ′ = 11.
This way we have two points with known coordinates and tangent slopes.
The first line is 𝑦 = 1 ∙ (𝑥 − 1) − 1 = 𝒙 − 𝟐, the second is 𝑦 = 11 ∙ (𝑥 − 1) − 3 = 𝟏𝟏𝒙 − 𝟏𝟒.

4. 𝑦 = (2 − 5𝑥 2 )4 (3𝑥 2 − 8)5 ...

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