Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 35537 by Raj Singh last updated on 20/May/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 20/May/18
	$f(x)=y=x^2 f(1)=1^2 =1 and f(2)=2^2 =4 slope m=((f(2)−f(1))/(2−1))=3 let (α,α^2 ) be the point ,slope of tangeny at(α,α^2 {((d(x^2 ))/dx)}_((α,α^2 )) =2α so 2α=3 α=3/2 so the cooridinate of the required poit is ((3/2),(9/4))$
	$f (x) = y = x^{2}$ $f (1) = 1^{2} = 1 a n d f (2) = 2^{2} = 4$ $s l o p e m = \frac{f (2) - f (1)}{2 - 1} = 3$ $l e t (α, α^{2}) b e t h e p o i n t, s l o p e o f t a n g e n y a t (α, α^{2}$ ${\frac{d (x^{2})}{d x}}_{(α, α^{2})} = 2 α$ $s o 2 α = 3 α = 3 / 2$ $s o t h e c o o r i d i n a t e o f t h e r e q u i r e d p o i t i s$ $(\frac{3}{2}, \frac{9}{4})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum