Matematika

Pertanyaan

Kak tolong bantuu pliss bua besok
Kak tolong bantuu pliss bua besok

1 Jawaban

  • Soal No. 1

    [tex]f(x)=x^2+4 \\ \\ g(x)= \frac{2}{ \sqrt{x} } \\ \\ gof(x)= \frac{2}{ \sqrt{x^2+4} } \\ \\ gof(x)= \frac{2}{ \sqrt{x^2+4} } \times \frac{\sqrt{x^2+4} }{ \sqrt{x^2+4} } \\ \\ gof(x)= \frac{2\sqrt{x^2+4}}{ x^2+4} [/tex]

    Soal No. 2
    [tex]f(x)= \frac{x}{x-1} \\ \\ g(x)= \sqrt{1+x^2} \\ \\ \\ fog(x)= \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2})-1} \\ \\ fog(x)= \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2})-1} \times \frac{(\sqrt{1+x^2})+1}{(\sqrt{1+x^2})+1} \\ \\ fog(x)= \frac{(\sqrt{1+x^2}) \times ((\sqrt{1+x^2})+1) }{(\sqrt{1+x^2})^2-(1^2)} \\ \\ fog(x)= \frac{(\sqrt{1+x^2})^2+ (\sqrt{1+x^2}) }{1+x^2-1} \\ \\ fog(x)= \frac{1+x^2+ (\sqrt{1+x^2}) }{x^2}[/tex]

    [tex]gof(x)= \sqrt{1+(\frac{x}{x-1})^2} \\ \\gof(x)= \sqrt{1+\frac{x^2}{(x-1)^2}}\\ \\gof(x)= \sqrt{1+\frac{x^2}{(x^2-2x+1)}} \\ \\gof(x)= \sqrt{\frac{(x^2-2x+1)}{(x^2-2x+1)}+\frac{x^2}{(x^2-2x+1)}}\\ \\gof(x)= \sqrt{\frac{x^2+x^2-2x+1}{x^2-2x+1}} \\ \\gof(x)= \sqrt{\frac{2x^2-2x+1}{x^2-2x+1}} [/tex]

    [tex]fog(2)=\frac{1+2^2+ (\sqrt{1+2^2}) }{2^2} \\ \\ fog(2)=\frac{5+(\sqrt{5})}{4}\\ \\ fog(2)=1\frac{1}{4}+ \frac{1}{4} \sqrt{5}[/tex]

    Soal No. 3

    [tex]f(x)=x^2-4x+5 \\ \\ g(x)=x+2 \\ \\ fog(x)=15 \\ \\ (x+2)^2-4(x+2)+5=15 \\ (x^2+4x+4)-4x-8+5=15 \\ x^2+1=15 \\ x^2=15-1 \\ x^2=14 \\ x= \sqrt{14} [/tex]

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