Ex 1.3, 9 - f(x) = 9x2 + 6x - 5. Show that f is invertible

Ex 1.3 , 9 - Chapter 1 Class 12 Relation and Functions - Part 2
Ex 1.3 , 9 - Chapter 1 Class 12 Relation and Functions - Part 3 Ex 1.3 , 9 - Chapter 1 Class 12 Relation and Functions - Part 4 Ex 1.3 , 9 - Chapter 1 Class 12 Relation and Functions - Part 5 Ex 1.3 , 9 - Chapter 1 Class 12 Relation and Functions - Part 6 Ex 1.3 , 9 - Chapter 1 Class 12 Relation and Functions - Part 7 Ex 1.3 , 9 - Chapter 1 Class 12 Relation and Functions - Part 8

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Ex 1.3, 9 Consider f: R+ → [-5, ∞) given by f(x) = 9x2 + 6x – 5. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = ((√(y +6)) − 1)/3 . f(x) = 9x2 + 6x – 5 f is invertible if it is one-one and onto Checking one-one f (x1) = 9(x1)2 + 6x1 – 5 f (x2) = 9(x2)2 + 6x2 – 5 Putting f (x1) = f (x2) 9(x1)2 + 6x1 – 5 = 9(x2)2 + 6x2 – 5 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = – 5 + 5 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = 0 9[(x1)2 – (x2)2 ]+ 6[x1 – x2] = 0 9[(x1 – x2) (x1 + x2) ]+ 6[x1 – x2] = 0 3(x1 – x2) [3(x1 + x2) + 2] = 0 (x1 – x2) [3x1 + 3x2 + 2] = 0/3 (x1 – x2) [3x1 + 3x2 + 2] = 0 (x1 – x2) = 0 ⇒ x1 = x2 (3x1 + 3x2 + 2) = 0 ⇒ 3x1 = – 3x2 – 2 Since f: R+ → [-5,∞ ) So x ∈ R+ i.e. x is always positive, Hence 3x1 = –3x2 – 2 is not true Hence, if f (x1) = f (x2) , then x1 = x2 ∴ f is one-one Check onto f(x) = 9x2 + 6x – 5 Let f(x) = y such that y ∈ [-5, ∞) Putting in equation y = 9x2 + 6x – 5 0 = 9x2 + 6x – 5 – y 9x2 + 6x – 5 – y = 0 9x2 + 6x – (5 + y) = 0 Comparing equation with ax2 + bx + c = 0 a = 9, b = 6 , c = – (5 + y) x = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐))/2𝑎 Putting values x = (− 6 ± √(6^2 − 4(9) (−(5 + 𝑦)) ))/2(9) x = (− 6 ± √(36 + 4(9)(5 + 𝑦) ))/18 = (− 6 ± √(36 + 36(5 + 𝑦)))/18 = (− 6 ± √(36(1 + (5 + 𝑦)) ))/18 = (− 6 ± √(36(6 + 𝑦) ))/18 = (− 6 ± √36 √((6 + 𝑦)))/18 = (− 6 ± √(6^2 ) √((6 + 𝑦)))/18 = (− 6 ± 6 √((6 + 𝑦)))/18 = 6[− 1 ± √((6 + 𝑦) )]/18 = (− 1 ± √((6 + 𝑦) ))/3 So, x = (− 1 − √((6 + 𝑦) ))/3 or (− 1 + √((6 + 𝑦) ))/3 As x ∈ R+ , i.e., x is a positive real number x cannot be equal to (−1 − √((6 + 𝑦) ))/3 Hence, x = (−𝟏 + √((𝟔 + 𝒚) ))/𝟑 Since f: R+ → [−5,∞ ) So y ∈ [−5,∞ ) i.e. y is greater than or equal to −5 i.e. y ≥ −5 y + 5 ≥ 0 Hence the value inside root is positive Hence √(𝑦+6) ≥ 0 x ≥ 0 Hence x is a real number which is greater than or equal to 0. ∴ x ∈ R+ Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((−𝟏 + √((𝟔 + 𝒚) ))/𝟑) = 〖9((−𝟏 + √((𝟔 + 𝒚) ))/𝟑)〗^2+6((−𝟏 + √((𝟔 + 𝒚) ))/𝟑)−5 = (−𝟏+√((𝟔 + 𝒚) ))^2+2(−𝟏+√((𝟔 + 𝒚) ))−5 = 1+(6+𝑦)−2√((𝟔 + 𝒚) )−2+2√((𝟔 + 𝒚) )−5 = 7+𝑦−7 = 𝑦 Thus, for every y ∈ [−5, ∞) , there exists x ∈ R+ such that f(x) = y Hence, f is onto Since f(x) is one-one and onto, So, f(x) is invertible And Inverse of x = 𝑓^(−1) (𝑦) = (−𝟏 + √((𝟔 + 𝒚) ))/𝟑

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.