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Diketahui begin mathsize 14px style straight f left parenthesis straight x right parenthesis equals straight k space left parenthesis straight x cubed minus 6 straight x squared plus 9 straight x right parenthesis comma space straight k greater than 0 end style, dan begin mathsize 14px style integral subscript 0 superscript straight a space straight f left parenthesis straight x right parenthesis equals 27 end style untuk (a,b) titik balik minimum. Nilai k adalah ....

  1. 9

  2. 8

  3. 6

  4. 4

  5. 3

A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

begin mathsize 14px style left parenthesis straight a comma straight b right parenthesis space titik space balik space minimum space maka space straight f apostrophe left parenthesis straight a right parenthesis equals 0  straight k space left parenthesis 3 straight a squared minus 12 straight a plus 9 right parenthesis equals 0  straight a squared minus 4 straight a plus 3 equals 0  left parenthesis straight a minus 3 right parenthesis left parenthesis straight a minus 1 right parenthesis equals 0  straight a equals 3 space atau space straight a equals 1    Karena space straight f left parenthesis straight a right parenthesis equals straight f left parenthesis 3 right parenthesis equals 0 space dan space straight f left parenthesis straight a right parenthesis equals straight f left parenthesis 1 right parenthesis equals 4 straight k comma space maka space titik space  minimum space yang space mungkin space hanyalah space ketika space straight a equals 3. space Jadi  integral subscript 0 superscript straight a space straight f left parenthesis straight x right parenthesis space dx equals 27  integral subscript 0 superscript 3 straight k space left parenthesis straight x cubed minus 6 straight x squared plus 9 straight x right parenthesis space dx equals 27  straight k integral subscript 0 superscript 3 straight x cubed minus 6 straight x squared plus 9 straight x space dx equals 27  straight k space open square brackets 1 fourth straight x to the power of 4 minus 2 straight x cubed plus 9 over 2 straight x squared close square brackets subscript 0 superscript 3 equals 27  straight k space open square brackets 81 over 4 minus 54 plus 81 over 2 close square brackets equals 27  27 over 4 straight k equals 27  straight k equals 4  end style

9

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Diketahui f(x)= k(x3-6x2 +9x), k>0 dan ∫0a​f(x)=27 untuk (a,b) titik balik minimum. Nilai k adalah ...

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