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Question

Find the slope of the tangent to the curve y=x3x+1 at the point whose x-coordinate is 1. Also find the equation of normal at the same point.

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Solution

y=x3x+1
dydxx=1=3x21=31=2
Thus the slope of the tangent at x=1 is 2.
slope of the normal at x=1 becomes 12
Thus, its equation can be written as y=x2+c
At x=1,y=11+1=1
Thus (1,1) lies on the normal and so, 1+12=c=32
Thus, equation of the normal is 2y+x=3

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