Find the equation of the tangent line to the curve y x23 1

Solution

slope of tangent of curve f(x) is given by f\'(x)

given, y = f(x) = -x^2 /3 + 1/(3x)

f(3/2) = -9/4*3 + 2/3*3 = -19/36

f\'(x) = -2x/3 + (1/3)(-1/x²)

f\'(x) = -(2x³+1)/3x²

f \'(3/2) = -31/27

slope, m = -31/27

y = mx + c

y = -19/36 at x = 3/2

-19/36 = (-31/27)*(3/2) + c

c = 43/36

eq of thangent is,

y = -(31/27)x + 43/36