Function Intervals

The function intervals are defined as the funcions that are appliable with intervals of the numbers which are applied in the variables that changes the functions.
Open Interval

This interval is common to denote open intervals in mathematics using parenthesis (  ). Thus the open interval discussed above will often be shown as (a, b),
Closed Interval

This type of interval are in the form of  [ ] closed both sides also.  For example consider that , -2< or equal to x < or equal 2.  This is represented by closed interval [2,2].  So open interval represents that the boundary value are not taken into the considerations, whereas closed interval means boundary value are taken into the considerations.

Intervals on the function:

Example 1:

Verify Lagrange’s law of the mean for the function f(x) = x3 on [−2,2]

Solution : f is a polynomial, hence continuous and differentiable on [− 2, 2].
f(2) = 23 = 8 ; f (−2) = (−2)3 = −8
f ′(x) = 3x2 ⇒ f ′(c) = 3c2
By law of the mean there exists an element c ∈ (− 2, 2) such that
f ′(c) = (f(b)−f(a)) / (b−a)
⇒ 3c2 = [ 8 − (−8) ] / 4 = 4
i.e., c2 = 4 / 3 ⇒ c = ± 2 / 3
The required ‘c’ in the law of mean are 2 / 3  and − 2/3  as both lie in [−2,2].

Example 2:

Suppose that the functions f(0) = − 3 and f ′(x) ≤ 5 for all values of x, how large can f(2) possibly be?
Solution :

Since by hypothesis f is differentiable, f is continuous everywhere. We can apply Lagrange’s Law of the mean on the interval [0,2]. There exist atleast one ‘c’∈(0, 2) such that
f(2) − f(0) = f ′(c) ( 2 − 0)
f(2) = f(0) + 2 f ′(c)
= −3 + 2 f ′(c)
Given that f ′(x) ≤ 5 for all x. In particular we know that f ′(c) ≤ 5.
Multiplying both sides of the inequality by 2, we have
2f ′(c) ≤ 10
f(2) = −3 + 2 f ′(c) ≤ −3 + 10 = 7
i.e., the largest possible value of f(2) is 7.

Example 3:

Determine the interval in the which curve f(x) =sinx ,  X  ∈ ( 0,2∏) only it be the concave downward

Solution :

f(x)  = Sinx

f'(x) = cosx

f''(x) = -sinx

In the interval (0,∏)  ,  -sinx <0

so f''(x) < 0 in the interval (0,∏)

∴f(x) = Sinx is concave downward in the interval (0,∏)

Practical applications on the function intervals

Example 1:

A cylindrical hole 4 mm in diameter and 12 mm deep in a metal block is rebored to increase the diameter to 4.12 mm. Estimate the amount of metal removed.
Solution :


The volume of cylindrical hole of radius x mm and depth 12 mm is given by

V = f(x) = 12 πx2
⇒ f ′(c) = 24πc.


To estimate f(2.06) − f(2) :
By law of mean,
f(2.06) − f(2) = 0.06 f ′(c)
= 0.06 (24 πc), 2 < c < 2.06

Take c = 2.01
f(2.06) − f(2) = 0.06 × 24 π × 2.01
= 2.89 π cubic mm.