# Ex 1.3 , 8

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

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Ex1.3 , 8 (Method 1) Consider f: R+ → [4, given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = y−4 , where R+ is the set of all non-negative real numbers. f(x) = x2 + 4 Step 1 Put f(x) = y y = x2 + 4 y – 4 = x2 x2 = y – 4 x = ±𝑦−4 Since f: R+ → [4,) x ∈ R+ , so x is positive Hence we cannot take x = −𝑦−4 ⇒ x = 𝑦−4 Let g(y) = 𝑦−4 where g: [4, → R+ Step 2: gof = g(f(x)) = g(x2 + 4) = x2 + 4 −4 = 𝑥2 = 𝑥12 × 2 = 𝑥1 = x Hence, gof(x) = IX = x Step 3: fog = f(g(y)) = f(𝑦−4 ) = (𝑦−4 )2 + 4 = (𝑦 −4)12 × 2+ 4 = (𝑦 −4)1 + 4 = y – 4 + 4 = y Hence, fog(y) = IY = y Since gof = IX and fog = IY, f is invertible & Inverse of f = g(y) = 𝑦−4 So, f-1 = 𝒚−𝟒 Ex1.3 , 8 (Method 2) Consider f: R+ → [4, given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = y−4 , where R+ is the set of all non-negative real numbers. f(x) = x2 + 4 f is invertible if f is one-one and onto Checking one-one f (x1) = (x1)2 + 4 f (x2) = (x2)2 + 4 Putting f (x1) = f (x2) ⇒ (x1)2 + 4 = (x2)2 + 4 ⇒ (x1)2 = (x2)2 ⇒ x1 = x2 or x1 = –x2 Since f: R+ → [4, So x ∈ R+ i.e. x is always positive, Hence x1 = –x2 is not true So, x1 = x2 ∴ f is one-one Check onto f(x) = x2 + 4 Let f(x) = y such that y ∈ [4, y = x2 + 4 y – 4 = x2 x2 = y – 4 x = ±𝑦−4 Since f: R+ → [4, x ∈ R+ , so x is positive Hence we cannot take x = −𝑦−4 ⇒ x = 𝑦−4 Since, y ∈ [4, i.e. y is greater than or equal to 4 i.e. y ≥ 4 y – 4 ≥ 0 ⇒Hence the value inside root is positive Hence 𝑦−4 ≥ 0 ⇒ x ≥ 0 Hence x is a real number which is greater than or equal to 0. ⇒ x ∈ R+ Hence, the function f is onto Since the function is one-one and onto ∴ It is invertible Calculating inverse For finding inverse, we put f(x) = y and find x in terms of y We have done that while proving onto x = 𝑦−4 Let g(y) = 𝑦−4 where g: [4, → R+ ∴ Inverse of f = g(y) = 𝑦−4 So, f-1 = 𝒚−𝟒

Chapter 1 Class 12 Relation and Functions

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .