Roboguru

Jika , maka f'(1) = ....

Jika begin mathsize 14px style limit as straight x rightwards arrow straight a of space fraction numerator straight f left parenthesis straight x cubed right parenthesis minus straight f left parenthesis straight a cubed right parenthesis over denominator straight x minus straight a end fraction equals negative 1 end style, maka f'(1) = ....

  1. -1

  2. begin inline style negative 1 third end style

  3. begin inline style 1 third end style

  4. 1

  5. 2

Jawaban:

begin mathsize 14px style Dengan space menggunakan space dalil space straight L ’ Hospital space diperoleh colon  limit as straight x rightwards arrow straight a of invisible function application fraction numerator straight f apostrophe open parentheses straight x cubed close parentheses straight space. straight space 3 straight x squared over denominator 1 end fraction equals negative 1  straight f apostrophe open parentheses straight a cubed close parentheses straight space. straight space 3 straight a squared equals negative 1  straight f apostrophe open parentheses straight a cubed close parentheses equals negative fraction numerator 1 over denominator 3 straight a squared end fraction    Karena space yang space diminta space adalah space straight f apostrophe left parenthesis 1 right parenthesis comma space maka  straight a cubed equals 1  straight a equals 1    Substitusikan space kembali space ke space persamaan space di space atas colon  straight f apostrophe space open parentheses 1 close parentheses equals negative fraction numerator 1 over denominator 3 open parentheses 1 close parentheses squared end fraction equals negative 1 third  end style

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