Date:22nd May 2020 Dr.

Sachin Kaushal (22206)

CONCAVITY AND CONVEXITY

Definitions : Let a function f be continuous on a closed interval [a, b] and differentiable in open
interval (a, b). Then
Y

(i) f is concave upwards on [a, b] if, throughout (a, b), the graph of f lies above the tangent
lines to f.
(ii) f is concave downwards on [a, b] if, throughout (a, b), the graph of lies below the tangent
lines to f.

Conditions for the concavity or convexity of the curve in the Upward Direction
d2 y d2 y
Concave upward 0 or Concave downward 0
dx 2 dx 2

POINT OF INFLEXION
Definition : A point on a curve, at which the curve changes from concavity to convexity or vice-
versa, is called a point of inflexion of the curve.

Test for Point of Inflexion

If f has an inflexion point at the number c in (a, b), then either f (c) = 0 or f  does not exist
at c. The converse is not necessarily true.
1. Find all numbers at which f (x) = 0
2. Use the test for concavity.
3. If the concavity changes, there is inflexion point; otherwise, there is not.
d2 y d2y
Thus a point is a point of inflexion, if  0 at this point and changes sign in passing
dx 2 dx 2
d3 y
through this point, i.e.,  0 at this point.
dx 3

Example 1: Find the range of values of x for which the curve y = 3x 5 40x3+ 3x20 is concave
upwards or downwards. Also find the points of inflexion on the curve.
Solution : The given curve is y = 3x5 40x3+ 3x  20 (1)

MTH125 Calculus
Date:22nd May 2020 Dr. Sachin Kaushal (22206)

dy
  15x 4  120x 2  3
dx
d2 y
2
 60x 3  240x  60x(x 2  4)
dx
(2)
d3 y
and  180x 2  240 (3)
dx 3
d2y
The curve (1) is concave upwards if >0
dx 2
i.e., if x(x2 4) > 0
or if x(x – 2)(x + 2) > 0
which is possible when either x ∈(2, 0) or x > 2 (4)
d2y
and the curve is concave downwards if <0
dx 2
i.e. if x(x – 2)(x + 2) < 0
i.e. when either x <2 or x ∈ (0, 2) (5)
Equations (4) and (5) give the range of values for the concavity or convexity of the curve in
the upward direction.
d2 y
The points of inflexion are given by 0
dx 2
Which gives x = 0, x = 2, x = 2.
d3 y
For all these values of x,  0 . Therefore these are the abscissa of the points of inflexion of the
dx 3
curve.
For x = 0, y = 20.
For x = 2, y = 3(2)5 40(2)3 + 3(2)  20
= 96 + 320  6 – 20 = 198
For x = 2, y = 3(2)5 40(2)3 + 3(2)  20
= 96  320 + 6  20 = 238
Hence the points of inflexion are (0, 20), (2, 198) and (2, 238).

Example 2 : Find the points of inflexion on the curve x = a(2sin), y = a(2cos).

Solution : Differentiating the given equation w.r.t. , we have
dx dy
 a(2  cos ) and  a sin 
d d

MTH125 Calculus
Date:22nd May 2020 Dr. Sachin Kaushal (22206)

dy
dy d sin 
  
dx dx 2  cos 
d
d 2 y (2  cos )cos   sin (sin ) d
and  .
dx 2 (2  cos ) 2 dx
2cos   cos 2   sin 2  1 2cos   1
 . 
(2  cos ) 2
a(2  cos ) a(2  cos )3
d2 y
For the points of inflexion, 0
dx 2
1 
i.e., 2 cos 1 = 0  cos =  cos
2 3

  = 2n , where n is any integer.
3
d2y
Obviously changes sign when  passes through each of the values given above. Hence
dx 2
each value of  corresponds to the point of inflexion on the curve. The co-ordinates of the points of
inflexion are given by
 2 3 3a
x  a  4n   ,y  .
 3 2  2

Example 3 : Find the points of inflexion of the curve y = (x – 2)6(x – 3)5.

Solution : The given curve is y = (x – 2)6(x – 3)5
dy
  (x  2)6 .5(x  3) 4  6(x  2)5 (x  3)5
dx
= (x – 2)5(x – 3)4[5(x – 2) + 6(x – 3)]
= (x – 2)5(x – 3)4(11x – 28)
d2 y
2
 (x  2)5 (x  3)4 .11  (x  2)5 (11x  28).4(x  3)3  (x  3) 4 (11x  28).5(x  2)4
dx
= (x – 2)4(x – 3)3[11(x – 2)(x – 3) + 4(x – 2) (11x – 28) + 5(x – 3)(11x – 28)]
= (x – 2)4(x – 3)3[110x2 560x + 710]
= 10(x – 2)4(x – 3)3(11x2 – 56x + 71)
 28  3   28  3 
 10(x  2)4(x  3)3  x   x  
 11   11 

d2 y
At the points of inflexion 0
dx 2
MTH125 Calculus
Date:22nd May 2020 Dr. Sachin Kaushal (22206)

28  3
 x  2,3,
11
d2y
At x = 2, does not change sign as the corresponding factor (x – 2)4 remains +ve whether x < 2
dx 2
or x > 2.
 x = 2 does not correspond to the point of inflexion.
d2y 28  3
Also 2
changes sign at x = 3 and x 
dx 11
Thus these correspond to the abscissa of the points of inflexion of the given curve.

Example 4 : Find the points of inflexion on the curve x2y = a2(x – y).
Solution : The given curve is x2y = a2(x – y)
a 2x
or y 2 (1)
x  a2
dy  (x 2  a 2 )  x.2x  a 2 (a 2  x 2 )
  a2  
dx  (x 2
 a 2 2
)  (x  a )
2 2 2

d2 y 2  (x  a ).( 2x)  (a  x ).2(x  a ).2x 

2 2 2 2 2 2
and  a  
dx 2  (x 2  a 2 )4 
2a 2 x(x 2  3a 2 )

(x 2  a 2 )3

d2 y
At the points of inflexion, 0
dx 2
 2a2 x(x2 3a2) = 0
 x = 0,  3a .
d2y
At x = 0, changes sign from +ve to ve and so there is a point of inflexion.
dx 2
d2y
Also changes sign from ve to +ve at x  3a , and thus there is a point of inflexion.
dx 2
d2y
At x   3a , changes sign from +ve to ve and so there is a point of inflexion.
dx 2
 3a 
Hence the points of inflexion are (0, 0),   3a,  .
 4 


MTH125 Calculus
Date:22nd May 2020 Dr. Sachin Kaushal (22206)

Exercise
1. Find the points of inflexion on the following curves
(i) r(2 – 1) = a2 (ii) r2 = a2.
2. Show that points of inflexion on the curve r = bn are given by r = b[n(n+1)]n/2.
3. Find the range of values of x for which the curve y = x 4 6x3 + 12x2 + 5x + 7 is concave or
convex upwards. Determine also the points of inflexion.
4. Find the points of inflexion for the following curves :
(i) y(a2 + x2) = x3 (ii) y = 3x4 4x3 + 1
5. Find the points of inflexion of the curve x = (y – 1)(y – 2)(y – 3).
6. Show that the points of inflexion of the curve y2 = (x – a)2(x – b) lies on the line 3x + a = 4b.
8. Prove that the curve y = log x is everywhere convex upwards.
Answers
 3a   1
1. (i)  , 3  (ii)  2a , 
 2   2
3. The curve is concave upwards in (, 1) and (2, ) and concave downwards in (1,2). The
points of inflexion are (1, 19) and (2, 33).
 3 3a   2 11 
4. (i) (0,0)   3a ,   (ii) (0,1) and  , 
 4   3 27 
5. (0, 2).

MTH125 Calculus