Ex 1.3, 4 - If f(x) = 4x - 3 / 6x - 4, show that fof(x) = x - Finding Inverse

  Ex 1.3 , 4 - Part 2
  Ex 1.3 , 4 - Part 3

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Ex1.3 , 4 If š‘“(š‘„)=﷐(4š‘„ āˆ’ 3)ļ·®6š‘„ āˆ’ 4ļ·Æ, š‘„ ≠ ﷐2ļ·®3ļ·Æ , show that š‘“š‘œš‘“(š‘„)=š‘„, for all š‘„ ≠ ﷐2ļ·®3ļ·Æ . What is the inverse of f? š‘“(š‘„)=﷐(4š‘„ āˆ’ 3)ļ·®6š‘„ āˆ’ 4ļ·Æ š‘“(š‘“ļ·š‘„ļ·Æ) = ﷐4š‘“(š‘„) āˆ’ 3ļ·®6š‘“(š‘„) āˆ’ 4ļ·Æ š‘“š‘œš‘“ļ·š‘„ļ·Æ = ﷐4﷐﷐4š‘„ āˆ’ 3ļ·®6š‘„ āˆ’ 4ļ·Æļ·Æ āˆ’ 3ļ·®6﷐﷐4š‘„ āˆ’ 3ļ·®6š‘„ āˆ’ 4ļ·Æļ·Æ āˆ’ 4ļ·Æ = ﷐﷐4﷐4š‘„ āˆ’ 3ļ·Æ āˆ’ 3﷐6š‘„ āˆ’ 4ļ·Æļ·®6š‘„ āˆ’ 4﷯﷮﷐6﷐4š‘„ āˆ’ 3ļ·Æ āˆ’ 4﷐6š‘„ āˆ’ 4ļ·Æļ·®6š‘„ āˆ’ 4ļ·Æļ·Æ = ﷐﷐16š‘„ āˆ’ 12 āˆ’ 18š‘„ +12ļ·®6š‘„ āˆ’ 4﷯﷮﷐24š‘„ āˆ’ 18 āˆ’ 24š‘„ +16ļ·®6š‘„ āˆ’ 4ļ·Æļ·Æ = ﷐16š‘„ āˆ’ 12 āˆ’ 18š‘„ +12ļ·®6š‘„ āˆ’ 4ļ·Æ Ɨ ﷐6š‘„ āˆ’ 4ļ·®24š‘„ āˆ’ 18 āˆ’ 24š‘„ + 16ļ·Æ = ﷐16š‘„ āˆ’ 12 āˆ’ 18š‘„ +12ļ·®24š‘„ āˆ’18 āˆ’24š‘„ +16ļ·Æ = ļ·āˆ’2š‘„ + 0ļ·®0 āˆ’ 2ļ·Æ = ļ·āˆ’2š‘„ļ·®āˆ’ 2ļ·Æ = x ∓ š‘“š‘œš‘“ļ·š‘„ļ·Æ = x Calculating inverse of f(x) š‘“(š‘„)=﷐(4š‘„ āˆ’ 3)ļ·®6š‘„ āˆ’ 4ļ·Æ Put f(x) = y y = ﷐(4š‘„ āˆ’ 3)ļ·®6š‘„ āˆ’ 4ļ·Æ y(6x – 4) = (4x – 3) 6xy – 4y = 4x – 3 6xy – 4x = 4y – 3 x(6y – 4) = 4y – 3 x = ﷐4š‘¦ āˆ’ 3ļ·®6š‘¦ āˆ’ 4ļ·Æ So, inverse of f = ﷐4š‘¦ āˆ’ 3ļ·®6š‘¦ āˆ’ 4ļ·Æ ∓ Let inverse of f = g (y) = ﷐4š‘¦ āˆ’ 3ļ·®6š‘¦ āˆ’ 4ļ·Æ g (y) = ﷐4š‘¦ āˆ’ 3ļ·®6š‘¦ āˆ’ 4ļ·Æ Replacing y with x g (x) = ﷐4š‘„ āˆ’ 3ļ·®6š‘„ āˆ’ 4ļ·Æ = f(x) Hence we can say inverse of f is f itself i.e. f -1 = f

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