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Hello! Can someone please teach me how to answer this problem? Determine the equation of line tangent to the graph of y^3=3x+2 at x=-1. The y^3 puzzles me a lot. :(
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help me also @ vicky007
take derivatives on both side you will finally get dy/dx = 1/y^2 now this is the slope of the tangent from x = -1 find the value of y by plugging x in 3y^= 3x+2 then find equation of tnagent in point slope form
Okay, I'll absorb this first. And thank you! I'll post a follow up if I don't understand something. :)
Hey again! Is the answer y=x-2?
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