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Increasing and Decreasing Functions at a Point

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Increasing and Decreasing Functions at a Point - Lesson Summary

Let f be a real valued function such that f: I → R.

Let x1 ∈ I and h > 0

Consider (x1 - h, x1 + h)

(1)f is said to be increasing at x1, if f is increasing in (x1 - h, x1 + h)

(2)f is said to be strictly increasing at x1, if f is strictly increasing in (x1 - h, x1 + h)

(3)f is said to be decreasing at x1, if f is decreasing in (x1 - h, x1 + h)

(4)f is said to be strictly decreasing at x1, if f is strictly decreasing in (x1 - h, x1 + h)


Theorem: Let f be continuous on [a, b] and differentiable on (a, b) Then:

(i) f is increasing in [a, b], if f ' (x) ≥ 0, ∀ x ∈ (a, b)

(ii) f is decreasing in [a, b], if f ' (x) ≤ 0, ∀ x ∈ (a, b)

(iii) f is a constant function in [a, b] if f ' (x) = 0, ∀ x ∈ (a, b)

Proof:

Let x1, x2 ∈ [a, b] such that x1 < x2

By the mean value theorem, there exists c ∈ (x1, x2) such that f(c) = f( x 2 ) - f( x 1 ) x 2 - x 1 ....(1)

(i)Let f(c) ≥ 0

⇒  f( x 2 ) - f( x 1 ) x 2 - x 1   ≥ 0

⇒ f( x 2 ) - f( x 1 )   ≥  0 (Since x1 < x2)

⇒  f( x 2 ) ≥ f( x 1 )

Or f( x 1 ) ≤ f( x 2 )

⇒ f is an increasing function.

(ii) Let f(c) ≤ 0

⇒   f( x 2 ) - f( x 1 ) x 2 - x 1   ≤ 0

f( x 2 ) - f( x 1 )  ≤ 0 (Since x1 < x2)

⇒ f( x 2 ) ≤ f( x 1 )

Or f( x 1 ) ≥ f( x 2 )

⇒ f is a decreasing function.

(iii)Let f(c) = 0

⇒  f( x 2 ) - f( x 1 ) x 2 - x 1   = 0

⇒ f( x 2 ) –   f( x 1 )  = 0

⇒ f( x 2 ) =  f( x 1 )  , ∀ x1, x2 ∈ [a, b]

⇒ f( x 1 ) =  f( x 2 ) 

Some results:

(I)f is strictly increasing in (a, b) if f ' (x) > 0, ∀ x ∈ [a, b]

(II)f is strictly decreasing in (a, b) if f ' (x) < 0, ∀ x ∈ [a, b]

(III)f is increasing or decreasing on R if it is increasing or decreasing in every interval of R

Function f is increasing in [a, b], if f ' (x) ≥ 0, ∀ x ∈ (a, b).

Function f is decreasing in [a, b], if f ' (x) ≤ 0, ∀ x ∈ (a, b).

Function f is a constant function in [a, b] if f ' (x) = 0, ∀ x ∈ (a, b).

Example :

Show that the function f(x) = x3 – 6x2 + 15x + 3 is strictly increasing on the set of real numbers.

Sol :

Given, f(x) = x3 – 6x2 + 15x + 3

Differentiating with respect to x, we get

f(x) = 3x2 - 12x + 15

     = 3(x2 - 4x + 5)

     = 3(x2 - 4x + 4 + 1)

     = 3((x - 2)2 + 1)

Since (x - 2)2 ≥ 1, ∀ x ∈ R

⇒ 3((x - 2)2 + 1) > 0, ∀ x ∈ R

⇒ f ' (x) > 0, ∀ x ∈ R

⇒ f(x) = x3 - 6x2 + 15x + 3 is strictly increasing on the set of real numbers.

Example :

Find the intervals in which the function, f, given by f(x) = x2 - 8x + 5 is:

(a) Strictly increasing

(b) Strictly decreasing

Sol :

Given f(x) = x2

Differentiating with respect to x, we get f(x) = 2x - 8

a) f is strictly increasing if f(x)

⇒ 2x - 8 > 0

⇒ 2x > 8

⇒ x > 4

⇒ x ∈ (4, ∞)

⇒ f is strictly increasing in (4, ∞)


b)  f  is strictly decreasing if f(x) < 0

⇒ 2x - 8 < 0

⇒ 2x < 8

⇒ x < 4

⇒ x ∈ (- ∞ , 4)

⇒ f is strictly decreasing in (- ∞ , 4)

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