Notes On Increasing and Decreasing Functions at a Point - CBSE Class 12 Maths
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) = $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ ....(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 x1 < x2) $\text{}$⇒ $\text{f(}{\text{x}}_{\text{2}}\text{)}\ge \text{f(}{\text{x}}_{\text{1}}\text{)}$ 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{≤ 0}$$\text{}$ (Since x1 < x2) ⇒ $\text{f(}{\text{x}}_{\text{2}}\text{)}\le \text{f(}{\text{x}}_{\text{1}}\text{)}$ 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 ⇒ = 0 ⇒ , ∀ x1, x2 ∈ [a, b] ⇒ 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)

#### 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) = $\frac{\text{f(}{\text{x}}_{\text{2}}\text{) - f(}{\text{x}}_{\text{1}}\text{)}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}$ ....(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 x1 < x2) $\text{}$⇒ $\text{f(}{\text{x}}_{\text{2}}\text{)}\ge \text{f(}{\text{x}}_{\text{1}}\text{)}$ 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{≤ 0}$$\text{}$ (Since x1 < x2) ⇒ $\text{f(}{\text{x}}_{\text{2}}\text{)}\le \text{f(}{\text{x}}_{\text{1}}\text{)}$ 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 ⇒ = 0 ⇒ , ∀ x1, x2 ∈ [a, b] ⇒ 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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