Question

A curve is such that dy/dx = 12/ (2x+1)squared . The point (1 , 1) lies on the curve .find the coordinates of the point at which the curve intersects the x axis

Answer 1

$\frac{}{}$ $\frac{dy}{dx }= \frac{12}{ (2x+1)^2}$

$dy= \frac{12{dx }}{ (2x+1)^2}$

$y= \int\frac{12{dx }}{ (2x+1)^2}=12\int\frac{{dx }}{ (2x+1)^2}=6\int\frac{{d (2x) }}{ (2x+1)^2}=\\= 6\int\frac{{d (2x+1) }}{ (2x+1)^2}$

$t=2x+1$

$\\ 6\int\frac{{d (2x+1) }}{ (2x+1)^2}= 6\int\frac{{d t }}{ t^2}= \frac{{-6 }}{ t}+C= \frac{-6 }{(2x+1) }+C$

$y= - \frac{6 }{(2x+1) }+C$

The point (1 , 1) lies on the curve:

$y(1)=1 \Rightarrow -\frac{6 }{(2*1+1) }+C=1 \Rightarrow \\C=1 -\frac{-6 }{(2*1+1) } \Rightarrow \\ \Rightarrow C=1 +2 \Rightarrow C=3$

$y= \frac{-6 }{2x+1 }+3 \\$

$y= \frac{6x-3 }{2x+1 }$

The coordinates of the point at which the curve intersects the x axis

$\frac{6x-3 }{2x+1 }=0 \Rightarrow {6x-3=0 } ( 2x+1 \neq 0 ) \Rightarrow x=\frac{1 }{2 }$ .

Answer 

The coordinates of the point at which the curve $y= \frac{6x-3 }{2x+1 }$ intersects the x axis:

$x=\frac{1 }{2 }$ .