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Consider the curve given by the equation (2y+1)^3 − 24x = −3.
(a) Show that dy/dx = 4/(2y+1)^2.

(b) Write an equation for the line tangent to the curve at the point (−1,−2).

(c) Evaluate d2y/dx2 at the point (−1,−2).

(d) The point ([tex]\frac{1}{6}[/tex],0) is on the curve. Find the value of [tex](y^{-1})'{(0})[/tex].

Sagot :

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