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).

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