How to Learn Calculus of One Variable, Volume 1 by J.D. Ghosh

By J.D. Ghosh

Intends to carry out essentially the most very important strategies affiliates with the theoretical facets. This publication can assist the coed to benefit the tools of Calculus and theoretical elements.

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Y = 1 − x + x−3 2x + 1 af Solution: Let f 1 x = 1 − x 2 2 Solved Examples Find the domain of the following: 1. y = tan–1 (2x + 1) Solution: 3 y is defined when − ∞ < 2 x + 1 < + ∞ 3 f1 (x) is defined when 1 − x ≥ 0 2 ⇔ x ≤ 1 ⇔ x ≤ 1 ⇔ −1 ≤ x ≤ 1 b a fg = ∴ D f1 x −1 , 1 = D1 (say) Function af Again, let f 2 x = FG x − 2 IJ ≥ 0 H x + 2K ⇔ x ≠ − 2 and b x − 2gb x + 2g ≥ 0 ⇔ x ≠ − 2 and a x − 2f b x − a−2fg ≥ x−3 2x + 1 af 3 f 1 x is defined when 3 f2 (x) is defined when 2 x + 1 ≠ 0 ⇔x≠− 1 2 R 1U ∴ D b f a x fg = R − S− V = D (say) T 2W F 1 I ∪ F − 1 , ∞I = −∞ , − H 2K H 2 K 1 1 Thus, D a y f = L−1 , − I ∪ F − , 1O = D MN 2 K H 2 PQ 2 2.

2. Domain of product of two functions f (x) and g (x) af af g (x) = dom f x ⋅ g x = dom (f (x)) ∩ dom (g (x)). 3. , = dom the domain of a rational function or the quotient function is the set of all real numbers with the exception of those real numbers for which the function in denominator becomes zero. Notes: 1. , f (x)). 2. (i) The statement “f (x) is defined for all x” signifies a f that f (x) is defined in the interval −∞ , ∞ . (ii) The statements “f (x) is defined in an interval finite or infinite” signifies that f (x) exists and is real for all real values of x belonging to the interval.

Y = log F 5x − x I > 0 ⇔ 5x – GH 4 JK 2 Solution: y is defined when x2 > 0 ⇔ x (5 – x) > 0 ⇔ (x – 0) (x – 5) < 0 ⇔ 0 < x a f ∴ D a y f = a0 , 5f F xI 2. : In the above two examples the functions in denominator are positive. This is why considerable function to be greater than zero is only the function in numerator. 3. y = log Fx GH x 2 2 I J + 4 x + 6K − 5x + 6 2 2 I >0 ⇔ J + 4x + 6K − 5x + 6 a f between 2 and 3 ⇔ x < 2 or x > 3 ⇔ x ∈ −∞ , 2 ∪ a3, + ∞f . : In the above example (3), the discriminant D = 16 – 4 × 1 × 6 = 16 – 24 = –ve for the function x2 + 4x + 6 in denominator which ⇒ x2 + 4x + 6 > 0.

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