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Pertanyaan

Diketahui fungsi f ( x ) = 3 x − 1 2 x − 1 ​ . Turunan pertama fungsi f ( x ) adalah f ( x ) . Nilai dari f ( 1 ) = ....

  1. long dash 3 space space

  2. 1 fourth

  3. 1 half

  4. 2 over 3

  5. 5 over 2 space

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N. Mustikowati

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

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Pembahasan

I n g a t factorial  F u n g s i space b e r b e n t u k space f left parenthesis x right parenthesis equals u over v  M a k a space t u r u n a n space p e r t a m a n y a space y a i t u colon space  f to the power of apostrophe left parenthesis x right parenthesis equals fraction numerator u to the power of apostrophe. v minus u. v to the power of apostrophe over denominator v squared end fraction  D i k e t a h u i colon  f left parenthesis x right parenthesis equals fraction numerator 2 x minus 1 over denominator 3 x minus 1 end fraction  M i s a l colon  u equals 2 x minus 1 left right arrow u to the power of apostrophe equals 2  v equals 3 x minus 1 left right arrow v to the power of apostrophe equals 3  M a k a colon  f apostrophe left parenthesis x right parenthesis equals fraction numerator 2 left parenthesis 3 x minus 1 right parenthesis minus left parenthesis 2 x minus 1 right parenthesis left parenthesis 3 right parenthesis over denominator left parenthesis 3 x minus 1 right parenthesis squared end fraction  f to the power of apostrophe left parenthesis x right parenthesis equals fraction numerator 6 x minus 2 minus 6 x plus 3 over denominator left parenthesis 3 x minus 1 right parenthesis squared end fraction  f to the power of apostrophe left parenthesis x right parenthesis equals fraction numerator 1 over denominator left parenthesis 3 x minus 1 right parenthesis squared end fraction  f to the power of apostrophe left parenthesis 1 right parenthesis equals fraction numerator 1 over denominator left parenthesis 3 left parenthesis 1 right parenthesis minus 1 right parenthesis squared end fraction  f to the power of apostrophe left parenthesis 1 right parenthesis equals 1 fourth  J a d i comma space n i l a i space d a r i space f to the power of apostrophe left parenthesis 1 right parenthesis space a d a l a h space 1 fourth.

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Hadi Muhammad Khairil

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