Applied Mathematics for Business and Economics

Lecture Note Differentiation

Suppose that f is a continuous function.

a. Findf′′

b. Find the hypercritical values of x. That is, find the values x=c where

+ fc′′()=0, or

+ f′′()c is undefined

c. Using the hypercritical values as endpoints of intervals, determine the

interval where

+ fx′′()> 0 and f is concave upward, and

+ fx′′()< 0 and f is concave downward.

d. Point of inflection occurs at those hypercritical values where f changes

concavity.

Example 1

For the function ()^32

3

5

2

fx x=− +x (a) determine the intervals on which f is

concave upward and the intervals on which it is concave downward, and (b) locate
any points of inflection.
Solution
a. We find the hypercritical values and then determine the concavity on the
related intervals.

()

()

()

32

2

3

5

2

33

63

fx x x

fx x x

fx x

= −+

′ =−

′′ =−

Now set fx′′()= 0 (There are no values where fx′′( )= 0 is undefined.)

1

630,

2

xx− ==

There are two intervals to be considered:

11

, and ,

22

⎛⎞⎛⎞

⎜⎟⎜−∞+

⎝⎠⎝

∞⎟

⎠

y y

x x

() )

Increasing, concave up

fx′′>>0,f x (^0) () ()
Increasing, concave down
′( fx′′′><0,f x 0
y y
x x

() ()

decreasing, concave up

fx′′′<>0,f x (^0) () ()
decreasing, concave down
fx′′′<<0,f x 0