Question-35537

Question Number 35537 by Raj Singh last updated on 20/May/18

Answered by tanmay.chaudhury50@gmail.com last updated on 20/May/18

f (x) = y = x^{2}

f (1) = 1^{2} = 1 a n d f (2) = 2^{2} = 4

s l o p e m = \frac{f (2) - f (1)}{2 - 1} = 3

l e t (α, α^{2}) b e t h e p o i n t, s l o p e o f t a n g e n y a t (α, α^{2}

{\frac{d (x^{2})}{d x}}_{(α, α^{2})} = 2 α

s o 2 α = 3 α = 3 / 2

s o t h e c o o r i d i n a t e o f t h e r e q u i r e d p o i t i s

(\frac{3}{2}, \frac{9}{4})