FINDING CONCAVITY AND INFLECTION POINTS FOR A FUNCTION

What is concave up and down ?

Concave upward :

If y = f(x) is a concave upward curve, then as x increases f'(x) either is of the same sign and increasing or changes sign from negative to positive.

In either case f'(x) is increasing and so f''(x) > 0 for a concave upward curve f''(x) > 0.

Concave downward :

An arc of a curve y = f(x) is called concave downward, then as x increases, f'(x) either is the same sign and decreasing or changes sign from positive to negative.

In either case f'(x) is decreasing and so f''(x) < 0, for a concave downward curve f''(x) < 0

concave-up-or-down

Procedure to find Concavity and Points of Inflection

To find concavity of a function y = f(x), we will follow the procedure given below.

Step 1 :

Find the first derivative f '(x). By equating the first derivative to 0, we will receive critical numbers.

Step 2 :

Find the second derivative, that is f ''(x). Equate the second derivative f ''(x) = 0.

Apply the values that we are deriving from the second derivative in the number line and decompose into intervals.

Step 3 :

From each interval, by applying one of the values in the second derivative, if we receive the second derivative.

  • f''(x) > 0, then the function f(x) is concave up in the interval.
  • f''(x) < 0, then the function f(x) is concave down in the interval.

Point of inflection :

A point of inflection is a point at which a curve is changing

concave upward to concave downward

or

concave downward to concave upward

A curve y = f(x) has one of its points x = c as an inflection point, if

i) f''(c) = 0 or is not defined and 

ii) f''(x) changes sign as x increases through x = c.

To find the point of inflection, we will follow the procedure given below.

Step 1 :

Equate the second derivative to 0, that is f ''(x) = 0

x = a, b

Step 2 :

(a, f(a)) and (b, f(b)) will be point of inflections.

For each problem, find the x - coordinates of all points of inflection, find all discontinuities, and find the open intervals where the function is concave up and concave down.

Problem 1 :

y = x3 - 3x2 + 4

Solution :

y = x3 - 3x2 + 4

Let y = f(x) 

f(x) = x3 - 3x2 + 4

f'(x)  = 3x2 - 6x

f''(x) = 6x - 6

f''(x) = 6(x - 1)

6(x - 1) = 0

x = 1

Decomposing into intervals

concave-up-or-downq1.png

Intervals

(-∞, 1)

f''(x) = 6(x - 1)

x = 0 ∈ (-∞, 1)

Sign of f ''(x)

f'' (0) = 6(0 - 1)

f'' (0) > 0

Concave up

(1, ∞)

x = 2 ∈ (-∞, 1)

f'' (2) = 6(2 - 1)

f'' (0) < 0

Concave down

Since the given is a polynomial function, it is continuous everywhere.

In f '' (x), changes its sign, there must be point of inflection.

When x = 1

f(1) =  13 - 3(1)2 + 4

= 1 - 3 + 4

= 2

So, the point of inflection is (1, 2).

Problem 2 :

y = x3 - 2x2 - 2

Solution :

y = x3 - 2x2 - 2

Let y = f(x) 

f(x) = x3 - 2x2 - 2

f'(x) = 3x2 - 4x

f''(x) = 6x - 4

6x - 4 = 0

x = 2/3

fun-is-concave-up-and-concave-down-s2

Intervals

(-∞, 2/3)

f''(x) = 2(3x - 2)

x = 0 ∈ (-∞, 2/3)

Sign of f ''(x)

f'' (0) = 2(-2)

f'' (0) < 0

Concave down

(2/3, ∞)

x = 2 ∈ (-∞, 1)

f'' (2) = 12 - 4

f'' (2) > 0

Concave up

Since the given is a polynomial function, it is continuous everywhere.

In f '' (x), changes its sign, there must be point of inflection.

When x = 2/3

f(2/3) =  (2/3)3 - 2(2/3)2 - 2

= 8/27 - 8/9 - 2

= (8 - 24 - 54)/27

= -70/27

So, the required point of inflection is (2/3, -70/27).

Problem 3 :

y = x4 + x3 - 3x2 + 1

Solution :

Let f(x) = x4 + x3 - 3x2 + 1

f '(x) = 4x3 + 3x2 - 6x + 0

f '' (x) = 12x2 + 6x - 6

f '' (x) = 0

6(2x2 + x - 1) = 0

6(2x - 1)(x + 2) = 0

x = 1/2 and x = -2

concave-up-or-downq2.png

(-∞, -2), (-2, 1/2) and (1/2, ∞)

Intervals

(-∞, -2)

f''(x) = (2x-1)(x+2)

x = -3 ∈ (-∞, -2)

Sign of f ''(x)

f'' (-3) = -7(-1)

= 7

f'' (-3) > 0

Concave up

(-2, 1/2)

x = 0 ∈ (-2, 1/2)

f'' (0) = -1(2)

= -2

f'' (0) < 0

Concave down

Since the given is a polynomial function, it is continuous everywhere.

In f '' (x), changes its sign, there must be point of inflection.

When x = -2

f(-2) =   (-2)4 + (-2)3 - 3(-2)2 + 1

= 16 + 8 - 12 +1

= 25 - 12

= 13

When x = 1/2

f(1/2) =   (1/2)4 + (1/2)3 - 3(1/2)2 + 1

= 1/16 + 1/8 - 3/4 + 1

= (1 + 2 - 12 + 16)/16

= 7/16

So, the point of inflection are (-2, 13) and (1/2, 7/16).

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