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Module 4 - CIN Problem.docx - Module 4 Discussion:...

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Module 4 – Discussion: Application CIN problemCIN #: a=2x2+y2=2x+2yFirst, Differentiate2x+2ydydx=2+2dydx2x−2=2dydx−2ydydx2x−2=(2−2)dydx2x−2=dydxSecond, Set dy/dx =0, tangent are parallel to the x-axis when the derivative is 02x−2=02x=2x=1Three, Substitute x in the original formula.(1)2+y2=2(1)+2y1+y2=2+2y1+y2−(2+2y)=2+2y−(2+2y)y2−2y−1=0Four, Solve for y using the quadratic equation.
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Unformatted text preview: y = −(− 2 ) ± √ (− 2 ) 2 − 4 ( 1 )(− 1 ) 2 ( 1 ) ¿ 2 ± √ 4 + 4 2 y = 2 ± √ 2 2 y = 1 + √ 2 ∨ y = 1 − √ 2 y ≈ 2.4242 ∨ y≈ − 0.4142 The horizontal tangent points are at points, (1, 2.41) and (1,-0.4141)....
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