Fine the point on the curve y=x/(1+x2) where the tangent to the curve has the 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)

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.
 
  • 82
Ya i got this answer...thanks sir
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Friend , I am not a Sir ! I am a plus two student just like you . Thanks for the reply !!
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ohh sorry..do u know the answer of that 3D question ie to find the point of intersection..
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The given lines are
Line 1 :    ( x – 1) /3   = (y – 1) /-1 , z+1 = 0
 
This can be written as
(x – 1)/3 = ( y -1 ) / -1 = ( z +1)/0 = λ          ……………(1)  [λ – a real number ]
 
Line 2 :  (x – 4)/2 = ( z + 1)/3 , y = 0
This can be written as
( x – 4)/2 = (y – 0)/0 = ( z+ 1)/3                          
 
Let P be the point of intersection.
As P lies on line  one , we can write the coordinates of point of intersection in terms of λ using (1) as follows
P ( 3 λ + 1 , - λ + 1 , -1)                  …………….(2)
 
This you get by equating each term of equation 1 separately with λ and then expressing x , y and z in terms of λ.
 
Let’s assume that the lines intersect each other. Then P should satisfy on Line 2 also. Let’s substitute for x , y and z in terms of λ as follows :
((3 λ + 1) – 4  )/2 =  (  ( - λ + 1) – 0 ) / 0 = ( ( -1) + 1) / 3
Let’s equate first two terms:
((3 λ + 1) – 4  )/2 =  (  ( - λ + 1) – 0 ) / 0
Ie , ( 3 λ – 3)/2  = (- λ + 1)/0
Cross multiplying,
- λ + 1 = 0
Or λ = 1
 
Now take last two terms
( ( - λ + 1 ) – 0 ) / 0 = ( -1 + 1)/3
Or -3 λ + 3 = 0
Or λ = 1
 
Similarly if you take the first and last terms and equate , you will still get λ = 1.
 
This consistency chows that the given lines intersect. Hence Proved.
 
Now , to find the point of intersection , substitute λ = 1 in (2).
We get Point as P ( 4 , 0 , -1).
 
Hope that helps.
 
 
  • 13
Can you do it with the help of second derivative test
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hi
 
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lol
 
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lol
  • -3
Yeah..but no idea
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Lol bro
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Hooool
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Himanchal Pradesh ki rajdhani konsi hai
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Please find this answer

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It's A
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Answer The question number 1 part second
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