Tentukan f (x) jika diketahui fungsi f berikut. b. f(x)=2x+1(x−5)(x+6)

Pertanyaan

Tentukan $\style{font-size:14px}{f\;\left(x\right)}$ jika diketahui fungsi $\style{font-size:14px}f$ berikut.

b. $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses x minus 5 close parentheses open parentheses x plus 6 close parentheses over denominator 2 x plus 1 end fraction end style$

T. Rizki

Master Teacher

Jawaban terverifikasi

Jawaban

diperoleh bahwa $f apostrophe left parenthesis x right parenthesis equals fraction numerator 2 x squared plus 2 x plus 61 over denominator open parentheses 2 x plus 1 close parentheses squared end fraction$ .

Pembahasan

Gunakan rumus turunan pembagian:

$y equals U over V space maka space y apostrophe equals fraction numerator U apostrophe times V space minus space V apostrophe times U over denominator V squared end fraction$

Menggunakan rumus turunan pembagian dapat diperoleh perhitungan sebagai berikut.

$begin mathsize 14px style f left parenthesis x right parenthesis equals fraction numerator left parenthesis x minus 5 right parenthesis left parenthesis x plus 6 right parenthesis over denominator 2 x plus 1 end fraction equals fraction numerator left parenthesis x squared plus x minus 30 right parenthesis over denominator 2 x plus 1 end fraction u equals x squared plus x minus 30 blank v equals 2 x plus 1 u apostrophe equals 2 x plus 1 blank v apostrophe equals 2 f apostrophe left parenthesis x right parenthesis equals fraction numerator u apostrophe v minus v apostrophe u over denominator v squared end fraction blank space space space space space space space equals fraction numerator left parenthesis 2 x plus 1 right parenthesis left parenthesis 2 x plus 1 right parenthesis minus 2 left parenthesis x squared plus x minus 30 right parenthesis over denominator left parenthesis 2 x plus 1 right parenthesis squared end fraction blank space space space space space space blank equals fraction numerator 4 x squared plus 4 x plus 1 minus 2 x squared minus 2 x plus 60 over denominator left parenthesis 2 x plus 1 right parenthesis squared end fraction space space space space space space space equals fraction numerator 2 x squared plus 2 x plus 61 over denominator left parenthesis 2 x plus 1 right parenthesis squared end fraction end style$

Dengan demikian, diperoleh bahwa $f apostrophe left parenthesis x right parenthesis equals fraction numerator 2 x squared plus 2 x plus 61 over denominator open parentheses 2 x plus 1 close parentheses squared end fraction$ .

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