Find the The Slopes of the tangent and the normal to the following curves at the indicated points :

y = x3 – x at x = 2

Given:


y = x3 – x at x = 2


First, we have to find of given function, f(x),i.e, to find the derivative of f(x)


(xn) = n.xn – 1


The Slope of the tangent is


y = x3 – x


(x3) + 3(x)


= 3.x3 – 1 – 1.x1 – 0


= 3x2 – 1


Since, x = 2


x = 2 = 3(2)2 – 1


x = 2 = (34) – 1


x = 2 = 12 – 1


x = 2 = 11


The Slope of the tangent at x = 2 is 11


The Slope of the normal =


The Slope of the normal =


The Slope of the normal =


1