CONCAVITY AND POINTS OF INFLECTION

Concave up :

The graph is said to be concave up (convex down) at a point if the tangent line to the graph at that point lies below the graph in the vicinity of the point.

Concave down :

It is said to be concave down (convex up) at a point if the tangent line lies above the graph in the vicinity of the point.

Relationship Between First Derivative and Concavity

Let f(x) be a function whose second derivative exists in an open interval I  =  (a, b). Then the function is said to be 

(i)  If f'(x) is strictly increasing on I, then the function is concave up on the open interval I.

(ii) If f'(x) is strictly decreasing on I, then the function is concave down on the open interval I.

Finding Concavity from Second Derivative

(i)  If f''(x) > 0 on an open interval I, then f(x) is concave up on I.

(i)  If f''(x) < 0 on an open interval I, then f(x) is concave down on I.

Point of Inflection

The points where the graph of the function changes from “concave up to concave down” or “concave down to concave up” are called the points of inflection of f (x) .

How to calculate point of inflection ?

(i) If f ′′(c) exists and f ′′(c) changes sign when passing through x = c , then the point

(c, f (c)) is a point of inflection of the graph of f .

(ii) If f ′′(c) exists at the point of inflection, then f ''(c) = 0 .

Inflection point and sharp point :

To determine the position of points of inflexion on the curve y = f (x) it is necessary to find the

points where f ′′(x) changes sign. For ‘smooth’ curves (no sharp corners), this may happen when either

(i) f ''(x) = 0 or

(ii) f ′′(x) does not exist at the point.

Problem 1 :

Find intervals of concavity and points of inflexion for the following functions:

(i) f (x) = x(x − 4)3

Solution :

f(x) = x(x − 4)3

u  =  x, v  =  (x − 4)3

u'  =  1 and v'  =  3(x-4)2

f'(x)  =  x[3(x-4)2] +  (x − 4)3 (1)

f'(x)  =  (x-4)2[3x + (x−4)]

f'(x)  =  (x-4)2(4x-4)

u  =  (x-4)2  v  =  (4x-4)

u'  =  2(x-4) and v'  =  4

f''(x)  =  (x-4)2(4)+(4x-4)2(x-4)

f''(x)  =  (x-4)[4(x-4)+2(4x-4)]

f''(x)  =  (x-4)[4x-16+8x-8]

f''(x)  =  (x-4)(12x-24)

f''(x)  =  12(x-4)(x-2)

f''(x)  =  0

x - 4  =  0 and x - 2  =  0

x  =  4 and x  =  2

At x  =  2, the curve changes its sign, it changes from concave up to concave down.

At x  =  4, the curve changes its sign, it changes from concave down to concave up.

Point of inflection :

At x  =  2

f (2) =  2(2 − 4)3

f (2) =  2(−2)3

f (2)  =  -16

At x  =  4

f (4)  =  2(4 − 4)3

f (4)  =  0

So, point of inflection are (2, -16) and (4, 0).

(ii)  y  =  sin x + cos x, 0 < x < 2π

Solution :

y  =  sin x + cos x

y'  =  cos x - sin x

y''  =  -sinx -cosx

y''  =   0

-sinx -cosx  =  0

-sinx  =  cosx

sinx/cosx  =  -1

tan x  =  -1

x  =  tan-1(-1)

x  =  3π/4, 7π/4

The intervals are (0, 3π/4) (3π/4, 7π/4) and (7π/4, 2π).

At x  =  3π/4, the curve changes its sign, it changes from concave down to concave up.

At x  =  7π/4, the curve changes its sign, it changes from concave up to concave down.

Point of inflection :

f(3π/4) 

=  sin (3π/4) + cos(3π/4)

=  -1/√2  + 1/√2 

=  0

f(7π/4) 

=  sin (7π/4) + cos(7π/4)

=  -1/√2  + 1/√2

=  0

So, points of inflection are (3π/4, 0) and (7π/4, 0).

Problem 3 :

f(x)  =  1/2 (ex-e-x)

Solution :

f(x)  =  1/2 (ex-e-x)

f'(x)  =  1/2 (ex+e-x)

f''(x)  =  1/2 (ex-e-x)

f''(x)  =  0

1/2 (ex-e-x)  =  0

ex  =  e-x

e2x  =  1

2x  =  0

x  =  0

So, the intervals are (-∞, 0) and (0, ∞).

Point of Inflection :

at x  =  0

f(0)  =  0

So, point of inflection is (0, 0).

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