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

Let x_{1} ∈ I and h > 0

Consider (x_{1} - h, x_{1} + h)

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

(2)f is said to be strictly increasing at x_{1}, if f is strictly increasing in (x_{1} - h, x_{1} + h)

(3)f is said to be decreasing at x_{1}, if f is decreasing in (x_{1} - h, x_{1} + h)

(4)f is said to be strictly decreasing at x_{1}, if f is strictly decreasing in (x_{1} - h, x_{1} + 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 x_{1}, x_{2} ∈ [a, b] such that x_{1} < x_{2}

By the mean value theorem, there exists c ∈ (x_{1}, x_{2}) such that f(c) = $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ ....(1)

(i)Let f(c) ≥ 0

⇒ $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ ≥ 0

⇒ $\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}$ ≥ 0 (Since x_{1} < x_{2})

$\text{}$⇒ $\text{f(}{\text{x}}_{\text{2}}\text{)}\ge \text{f(}{\text{x}}_{\text{1}}\text{)}$

Or $\text{f(}{\text{x}}_{\text{1}}\text{)}\le \text{f(}{\text{x}}_{\text{2}}\text{)}$

⇒ f is an increasing function.

(ii) Let f(c) ≤ 0

⇒ $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ ≤ 0

$\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}$$\text{\u2264 0}$$\text{}$ (Since x_{1} < x_{2})

⇒ $\text{f(}{\text{x}}_{\text{2}}\text{)}\le \text{f(}{\text{x}}_{\text{1}}\text{)}$

Or $\text{f(}{\text{x}}_{\text{1}}\text{)}\ge \text{f(}{\text{x}}_{\text{2}}\text{)}$

⇒ f is a decreasing function.

(iii)Let f(c) = 0

⇒ $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ = 0

⇒ $\text{f(}{\text{x}}_{\text{2}}\text{) \u2013}\text{f(}{\text{x}}_{\text{1}}\text{)}$ = 0

⇒ $\text{f(}{\text{x}}_{\text{2}}\text{) =}\text{f(}{\text{x}}_{\text{1}}\text{)}$ , ∀ x_{1}, x_{2} ∈ [a, b]

⇒ $\text{f(}{\text{x}}_{\text{1}}\text{) =}\text{f(}{\text{x}}_{\text{2}}\text{)}$

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) = x^{3} – 6x^{2} + 15x + 3 is strictly increasing on the set of real numbers.

**Sol** :

Given, f(x) = x^{3} – 6x^{2} + 15x + 3

Differentiating with respect to x, we get

f(x) = 3x^{2} - 12x + 15

= 3(x^{2} - 4x + 5)

= 3(x^{2} - 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) = x^{3} - 6x^{2} + 15x + 3 is strictly increasing on the set of real numbers.

**Example** :

Find the intervals in which the function, f, given by f(x) = x^{2} - 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)

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

Let x_{1} ∈ I and h > 0

Consider (x_{1} - h, x_{1} + h)

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

(2)f is said to be strictly increasing at x_{1}, if f is strictly increasing in (x_{1} - h, x_{1} + h)

(3)f is said to be decreasing at x_{1}, if f is decreasing in (x_{1} - h, x_{1} + h)

(4)f is said to be strictly decreasing at x_{1}, if f is strictly decreasing in (x_{1} - h, x_{1} + 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 x_{1}, x_{2} ∈ [a, b] such that x_{1} < x_{2}

By the mean value theorem, there exists c ∈ (x_{1}, x_{2}) such that f(c) = $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ ....(1)

(i)Let f(c) ≥ 0

⇒ $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ ≥ 0

⇒ $\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}$ ≥ 0 (Since x_{1} < x_{2})

$\text{}$⇒ $\text{f(}{\text{x}}_{\text{2}}\text{)}\ge \text{f(}{\text{x}}_{\text{1}}\text{)}$

Or $\text{f(}{\text{x}}_{\text{1}}\text{)}\le \text{f(}{\text{x}}_{\text{2}}\text{)}$

⇒ f is an increasing function.

(ii) Let f(c) ≤ 0

⇒ $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ ≤ 0

$\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}$$\text{\u2264 0}$$\text{}$ (Since x_{1} < x_{2})

⇒ $\text{f(}{\text{x}}_{\text{2}}\text{)}\le \text{f(}{\text{x}}_{\text{1}}\text{)}$

Or $\text{f(}{\text{x}}_{\text{1}}\text{)}\ge \text{f(}{\text{x}}_{\text{2}}\text{)}$

⇒ f is a decreasing function.

(iii)Let f(c) = 0

⇒ $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ = 0

⇒ $\text{f(}{\text{x}}_{\text{2}}\text{) \u2013}\text{f(}{\text{x}}_{\text{1}}\text{)}$ = 0

⇒ $\text{f(}{\text{x}}_{\text{2}}\text{) =}\text{f(}{\text{x}}_{\text{1}}\text{)}$ , ∀ x_{1}, x_{2} ∈ [a, b]

⇒ $\text{f(}{\text{x}}_{\text{1}}\text{) =}\text{f(}{\text{x}}_{\text{2}}\text{)}$

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) = x^{3} – 6x^{2} + 15x + 3 is strictly increasing on the set of real numbers.

**Sol** :

Given, f(x) = x^{3} – 6x^{2} + 15x + 3

Differentiating with respect to x, we get

f(x) = 3x^{2} - 12x + 15

= 3(x^{2} - 4x + 5)

= 3(x^{2} - 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) = x^{3} - 6x^{2} + 15x + 3 is strictly increasing on the set of real numbers.

**Example** :

Find the intervals in which the function, f, given by f(x) = x^{2} - 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)