find point on curve y=x/x^2+1 where tangent to curve has greatest slope
Here,
y = x / 1 + x2
Slope of the tangent to this curve at any point is given by
M = dy/dx = ( 1 + x2) – 2x(x)/( 1 + x2)2 (using quotient rule)
= 1- x2/ (1 + x2)2
This is what we need to maximize.
Let’s differentiate M w.r.t x using quotient rule.
dM/dx = -2x( 1 + x2)2 - 2( 1 + x2) 2x ( 1 – x2) / ( 1 + x2)4
= -2x( 1 + x2) [ 1 + x2 + 2 ( 1 – x2) ] / ( 1 + x2)4
= -2x( 1 + x2)( 3 – x2) / ( 1 + x2)4
At maxima or minima , dM/dx = 0
So we get -2x( 1 + x2)( 3 – x2) = 0
Which gives x = 0 , √3 , - √3
On doing the first derivative test , it can be seen that as the function crosses these turning points , at x = 0 , its value changes from +ve to –ve and at other two points , the value change from –ve to +ve . So 0 is the point of maxima and others are the point of minima .
So the x coordinate of the point at which curve has maximum slope = 0 .
Substituting x = 0 in the equation of curve , we get y = 0.
So the required point is ( 0 , 0 ) .
Hope that helps.
y = x / 1 + x2
Slope of the tangent to this curve at any point is given by
M = dy/dx = ( 1 + x2) – 2x(x)/( 1 + x2)2 (using quotient rule)
= 1- x2/ (1 + x2)2
This is what we need to maximize.
Let’s differentiate M w.r.t x using quotient rule.
dM/dx = -2x( 1 + x2)2 - 2( 1 + x2) 2x ( 1 – x2) / ( 1 + x2)4
= -2x( 1 + x2) [ 1 + x2 + 2 ( 1 – x2) ] / ( 1 + x2)4
= -2x( 1 + x2)( 3 – x2) / ( 1 + x2)4
At maxima or minima , dM/dx = 0
So we get -2x( 1 + x2)( 3 – x2) = 0
Which gives x = 0 , √3 , - √3
On doing the first derivative test , it can be seen that as the function crosses these turning points , at x = 0 , its value changes from +ve to –ve and at other two points , the value change from –ve to +ve . So 0 is the point of maxima and others are the point of minima .
So the x coordinate of the point at which curve has maximum slope = 0 .
Substituting x = 0 in the equation of curve , we get y = 0.
So the required point is ( 0 , 0 ) .
Hope that helps.