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.

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 :

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 :

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

Problem 3 :

f(x)  =  1/2 (e^x-e^-x)

Solution :

f(x)  =  1/2 (e^x-e^-x)

f'(x)  =  1/2 (e^x+e^-x)

f''(x)  =  1/2 (e^x-e^-x)

f''(x)  =  0

1/2 (e^x-e^-x)  =  0

e^x  =  e^-x

e^2x  =  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)

Problem 4 :

What are the coordinates of the points of inflection for the graph of f(x) = sin²x on the closed interval [0, π] ?

a) x = 3π/4 only 

b)  x = π/4, x = π/2 and x = 3π/4

c)  x = π/4 and x = 3π/4

d) x = π/2             e) x = π/4

Solution :

f(x) = sin²x

f'(x) = 2 sin x cos x

= sin 2x

To find point of inflection, we have to equate the second derivative to 0.

f'(x) = sin 2x

f''(x) = cos 2x (2)

f''(x) = 2 cos 2x

f''(x) = 0

2 cos 2x = 0

2x = cos^-1(0)

2x = π/2, 3π/2

x = π/4, 3π/4

So, in the interval  [0, π] the given function has point of inflection at the point x = π/4 and 3π/4. So, option c is correct.

Problem 5 :

The function g is defined by the equation

g(x) = 6x⁵ - 10x³

Determine the values of x for which the graph of function g is concaved upwards

a) x > 1/2     b) -√2/2 < x < 0 or x > √2/2

c)  -1/2 < x < 0 or x > 1/2      d) -1/2 < x < 1/2

e) -√2/2 < x < √2/2

Solution :

g(x) = 6x⁵ - 10x³

g'(x) = 30x⁴ - 30x²

g''(x) = 120x³ - 60x

g''(x) = 0

120x³ - 60x = 0

60x(2x² - 1) = 0

x = 0 and 2x² - 1 = 0

 2x² = 1

x² = 1/2

x = -1/√2 and x = 1/√2

-1/2 < x < 0

When x = -0.4 

f''(-0.4) = 60(-0.4)(2(-0.4)² - 1)

= -24(0.32 - 1)

= -24(-0.68)

= + > 0

Concave up.

0 < x < 1/2

When x = 0.4 

f''(0.4) = 60(0.4)(2(0.4)² - 1)

= 24(0.32 - 1)

= 24(-0.68)

= - < 0

Concave down

x > 1/2

When x = 1

f''(1) = 60(1)(2(1)² - 1)

= 60(2 - 1)

= + > 0

Concave up

The function is concave up in the intervals

-1/2 < x < 0 or x > 1/2

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