Pertanyaan

Diketahui f ( x ) = x 1 + x , jika f ( x ) turunan f ( x ) , f ( 4 ) = 9 . Tentukan fungsi nya!

Diketahui $f (x) = \frac{1}{x} + x$ , jika $f (x)$ turunan $f (x)$ , $f (4) = 9$ . Tentukan fungsi f open parentheses x close parentheses nya!

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Habis dalam

Klaim

L. Rante

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Pembahasan

, dengan melakukan manipulasi aljabar di peroleh merupakan anti-turunan dari , yaitu Jadi, . Karena , maka Substitusi nilai ke , diperoleh .

$f apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator square root of x end fraction plus square root of x$ , dengan melakukan manipulasi aljabar di peroleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell f apostrophe open parentheses x close parentheses end cell equals cell fraction numerator 1 over denominator square root of x end fraction plus square root of x end cell row blank equals cell 1 over x to the power of begin display style 1 half end style end exponent plus x to the power of 1 half end exponent end cell row cell f apostrophe open parentheses x close parentheses end cell equals cell x to the power of negative 1 half end exponent plus x to the power of 1 half end exponent end cell end table$

f open parentheses x close parentheses merupakan anti-turunan dari f apostrophe open parentheses x close parentheses , yaitu

$table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell integral f apostrophe open parentheses x close parentheses d x end cell row blank equals cell integral open parentheses x to the power of negative 1 half end exponent plus x to the power of 1 half end exponent close parentheses d x end cell row blank equals cell fraction numerator 1 over denominator begin display style 1 half end style end fraction x to the power of 1 half end exponent plus fraction numerator 1 over denominator begin display style 3 over 2 end style end fraction x to the power of 3 over 2 end exponent plus C end cell row blank equals cell 2 x to the power of 1 half end exponent plus 2 over 3 x to the power of 3 over 2 end exponent plus C end cell row blank equals cell 2 x to the power of 1 half end exponent plus 2 over 3 x times x to the power of 1 half end exponent plus C end cell row blank equals cell 2 square root of x plus 2 over 3 x square root of x plus C end cell end table$

Jadi, f open parentheses x close parentheses equals 2 square root of x plus 2 over 3 x square root of x plus C . Karena f open parentheses 4 close parentheses equals 9 , maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell 2 square root of x plus 2 over 3 x square root of x plus C end cell row cell f open parentheses 4 close parentheses end cell equals cell 2 square root of 4 plus 2 over 3 open parentheses 4 close parentheses square root of 4 plus C end cell row 9 equals cell 4 plus 16 over 3 plus C end cell row C equals cell 9 minus 4 minus 16 over 3 end cell row C equals cell 5 minus 16 over 3 end cell row C equals cell fraction numerator 15 minus 16 over denominator 3 end fraction end cell row C equals cell negative 1 third end cell end table$

Substitusi nilai C equals negative 1 third ke f open parentheses x close parentheses equals 2 square root of x plus 2 over 3 x square root of x plus C , diperoleh .