Fungsi  dan , jika f−1 adalah infers dari f, maka . Maka nilai dari k+p=....

Pertanyaan

Fungsi begin mathsize 14px style f open parentheses x close parentheses equals 2 to the power of 3 x minus 1 end exponent end style dan begin mathsize 14px style g open parentheses x close parentheses equals 4 open parentheses x plus 2 close parentheses cubed end style, jika begin mathsize 14px style f to the power of negative 1 end exponent end style adalah infers dari f, maka begin mathsize 14px style open parentheses f to the power of negative 1 end exponent o g close parentheses open parentheses x close parentheses equals log presubscript presuperscript 2 invisible function application k open parentheses x plus p close parentheses end style. Maka nilai dari begin mathsize 14px style k plus p equals.... end style  

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M. Robo

Master Teacher

Jawaban terverifikasi

Jawaban

k + p = 4

Pembahasan

Infers dari begin mathsize 14px style f open parentheses x close parentheses end style :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 2 to the power of 3 x minus 1 end exponent end cell row cell log presubscript blank presuperscript 2 invisible function application y end cell equals cell 3 x minus 1 end cell row cell 1 third left parenthesis log presubscript blank presuperscript 2 invisible function application y plus 1 right parenthesis end cell equals x row cell f to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell 1 third left parenthesis log presubscript blank presuperscript 2 invisible function application x plus 1 right parenthesis end cell end table end style   

Maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f to the power of negative 1 end exponent o g close parentheses open parentheses x close parentheses end cell equals cell f to the power of negative 1 end exponent open parentheses g open parentheses x close parentheses close parentheses end cell row blank equals cell 1 third open parentheses log presubscript blank presuperscript 2 invisible function application 4 open parentheses x plus 2 close parentheses cubed plus 1 close parentheses end cell row blank equals cell 1 third open parentheses log presubscript blank presuperscript 2 invisible function application 4 plus log presubscript blank presuperscript 2 invisible function application open parentheses x plus 2 close parentheses cubed plus 1 close parentheses end cell row blank equals cell 1 third open parentheses 2 plus 3 log presubscript blank presuperscript 2 invisible function application open parentheses x plus 2 close parentheses plus 1 close parentheses end cell row blank equals cell 1 third open parentheses 3 plus 3 log presubscript blank presuperscript 2 invisible function application open parentheses x plus 2 close parentheses close parentheses end cell row blank equals cell 1 plus log presubscript blank presuperscript 2 invisible function application open parentheses x plus 2 close parentheses end cell row blank equals cell log presubscript blank presuperscript 2 invisible function application 2 plus log presubscript blank presuperscript 2 invisible function application open parentheses x plus 2 close parentheses end cell row blank equals cell log presubscript blank presuperscript 2 invisible function application 2 open parentheses x plus 2 close parentheses end cell end table end style   

 

Samakan dengan soal

begin mathsize 14px style log presubscript blank presuperscript 2 invisible function application k open parentheses x plus p close parentheses equals log presubscript blank presuperscript 2 invisible function application 2 open parentheses x plus 2 close parentheses end style   

Didapat k = 2 dan p = 2.
Sehingga k + p = 4

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