# Ex 1.3 , 9

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 1.3 , 9 (Method 1) 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 − 13 . f(x) = 9x2 + 6x – 5 Step 1: Let f(x) = y y = 9x2 + 6x – 5 0 = 9x2 + 6x – 5 – y 9x2 + 6x – 5 – y = 0 9x2 + 6x – (5 + y) = 0 x = −𝑏 ± 𝑏2 − 4𝑎𝑐2𝑎 Putting values x = − 6 ± 62 − 49 −5 + 𝑦29 x = − 6 ± 36 + 495 + 𝑦18 = − 6 ± 36 + 36(5 + 𝑦)18 = − 6 ± 361 + 5 + 𝑦18 = − 6 ± 366 + 𝑦18 = − 6 ± 36(6 + 𝑦)18 = − 6 ± 62(6 + 𝑦)18 = − 6 ± 6 (6 + 𝑦)18 = 6− 1 ± 6 + 𝑦18 = − 1 ± 6 + 𝑦3 So, x = − 1 − 6 + 𝑦3 or − 1 + 6 + 𝑦3 As x ∈ R+ , i.e., x is a positive real number x cannot be equal to −1 − 6 + 𝑦 3 Hence, x = −𝟏 + 𝟔 + 𝒚𝟑 Let g(y) = −𝟏 + 𝟔 + 𝒚𝟑 where g: [–5, → R+ Step 2: gof = g(f(x)) = g(9x2 + 6x – 5) = −1 + 6 + 9𝑥2 + 6𝑥 − 53 = −1 + 9𝑥2 + 6𝑥 + 13 = −1 + (3𝑥)2+ 12 +23𝑥 × 23 = −1 + 3𝑥 + 123 = −1 + 3𝑥 + 13 = 3𝑥3 = x Hence, gof = x = IR+ Step 3: fog = f(g(x)) = f(−1 + 6 + 𝑦3) = 9−1 + 6 + 𝑦32 + 6−1 + 6 + 𝑦3 – 5 = 9−1 + 6 + 𝑦29 + 2−1 + 6 + 𝑦 – 5 = −1 + 6 + 𝑦2– 2 + 26 + 𝑦 – 5 = (–1)2 + 6 + 𝑦2– 26 + 𝑦 – 2 + 26 + 𝑦 – 5 = 1 + (6 + y) – 2 – 5 = 7 – 7 + y = y Hence, fog = y = I[–5, Since, gof = IR+ & fog = I[–5, f is invertible, and inverse of f = g(y) = −1 + 6 + 𝑦3 So, f–1 = −𝟏 + 𝟔 + 𝒚𝟑 Ex 1.3 , 9 (Method 2) 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 − 13 . 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 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] = 03 (x1 – x2) [3x1 + 3x2 + 2] = 0 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 x = −𝑏 ± 𝑏2 − 4𝑎𝑐2𝑎 Putting values x = − 6 ± 62 − 49 −5 + 𝑦29 x = − 6 ± 36 + 495 + 𝑦18 = − 6 ± 36 + 36(5 + 𝑦)18 = − 6 ± 361 + 5 + 𝑦18 = − 6 ± 366 + 𝑦18 = − 6 ± 36(6 + 𝑦)18 = − 6 ± 62(6 + 𝑦)18 = − 6 ± 6 (6 + 𝑦)18 = 6− 1 ± 6 + 𝑦18 = − 1 ± 6 + 𝑦3 So, x = − 1 − 6 + 𝑦3 or − 1 + 6 + 𝑦3 As x ∈ R+ , i.e., x is a positive real number x cannot be equal to −1 − 6 + 𝑦 3 Hence, x = −𝟏 + 𝟔 + 𝒚𝟑 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+ 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 = −1 + 6 + 𝑦3 Let g(y) = −1 + 6 + 𝑦3 where g: [–5, → R+ So, inverse of f = f–1 = −𝟏 + 𝟔 + 𝒚𝟑

Ex 1.2, 5
Important

Ex 1.2 , 10 Important

Example 23 Important

Example 25 Important

Ex 1.3, 3 Important

Ex 1.3 , 6 Important

Ex 1.3 , 8 Important

Ex 1.3 , 9 Important You are here

Ex 1.3 , 13 Important

Ex 1.3 , 14 Important

Ex 1.4, 11 Important

Misc 3 Important

Misc. 4 Important

Misc 14 Important

Misc 18 Important

Class 12

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