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## Homework Statement

Consider:

[tex]x^3+y^3+2xy=4[/tex], y=1 when x=1

a.) Find the equation of the tangent line to the curve when x=1.

b.) Find y'' at x=1.

c.) Is the graph of y=f(x) concave up or concave down near x=1?

## Homework Equations

Any derivative rules...

## The Attempt at a Solution

__For Part a:__[tex]y'=\frac{-2y-3x^2}{3y^2+2x}[/tex]

After substituting x=1 and y=1:

[tex]y'=-1[/tex]

For y=mx+B, B=2

[tex]y=-x+2[/tex]

__For Part b:__[tex]y''=\frac{vu'-uv'}{v^2}[/tex]

[tex]u=-2y-3x^2[/tex]

[tex]u'=-2y'-6x[/tex]

[tex]v=3y^2+2x[/tex]

[tex]v'=6yy'+2[/tex]

[tex]y''=\frac{-(3y^2+2x)(2y'+6x)+(6yy'+2)(2y+3x^2)}{(3y^2+2x)^2}[/tex]

After substituting x=1, y=1, and y'=-1:

[tex]y''=-1.6[/tex]

__For Part c:__Would it be concave down because I just tested y'' with x=1?

I'm mostly unsure about my steps in part b.) and c.), and I want to make sure I'm doing it correctly. Thanks for your help.