Tolong bantu saya,saya tidak mengerti sama sekali.terima kasih.... Question from @Aprisillaj

June 2019 1 87 Report

Tolong bantu saya,saya tidak mengerti sama sekali.terima kasih

Verified answer

Soal a.
(f o g)(x) = 2(3x - 4)² + (3x - 4)
(f o g)(x) = 2(9x² - 24x + 16) + 3x - 4
(f o g)(x) = 18x² - 48x + 32 + 3x - 4
(f o g)(x) = 18x² - 45x + 28

$(h\ o\ (f\ o\ g))(x)= \frac{2(18x^2-45x+28)-4}{-4-(18x^2-45x+28)} \\ \\ (h\ o\ (f\ o\ g))(x)= \frac{36x^2-90x+56-4}{-18x^2+45x-28-4}\\ \\ (h\ o\ (f\ o\ g))(x)= \frac{36x^2-90x+52}{-18x^2+45x-32}\\ \\ (h\ o\ (f\ o\ g))(x)= -\frac{2(18x^2-45x+26)}{18x^2-45x+32}$

Soal b.
(g o f)(x) = 3(2x² + x) - 4
(g o f)(x) = 6x² + 3x - 4

$(h\ o\ g\ o\ f)(x)= \frac{2(6x^2+3x-4)-4}{-4-(6x^2+3x-4)} \\ \\ (h\ o\ g\ o\ f)(x)= \frac{12x^2+6x-8-4}{-4-6x^2-3x+4}\\ \\ (h\ o\ g\ o\ f)(x)= \frac{12x^2+6x-12}{-6x^2-3x}\\ \\ (h\ o\ g\ o\ f)(x)= \frac{6(2x^2+x-2)}{-3x(2x-1)}\\ \\ (h\ o\ g\ o\ f)(x)= -\frac{2(2x^2+x-2)}{x(2x-1)}$

Soal c

$(f\ o\ h)(x)=2( \frac{2x+4}{-4-x})^2+( \frac{2x+4}{-4-x}) \\ \\(f\ o\ h)(x)=2( \frac{4x^2+16x+16}{(-4-x)^2})+( \frac{2x+4}{-4-x})\\ \\(f\ o\ h)(x)= \frac{8x^2+32x+32}{(-4-x)^2}+( \frac{2x+4}{-4-x})\\ \\(f\ o\ h)(x)= \frac{(8x^2+32x+32)(-4-x)+(2x+4)(-4-x)^2}{(-4-x)^2(-4-x)}\\ \\(f\ o\ h)(x)= \frac{-8x^3-64x^2-160x-128+2x^3+20x^2+64x+64}{-x^3-12x^2-48x-64}\\ \\(f\ o\ h)(x)= \frac{-6x^3-44x^2-96x-64}{-x^3-12x^2-48x-64}\\ \\(f\ o\ h)(x)= \frac{-(6x^3+44x^2+96x+64)}{-(x^3+12x^2+48x+64)}$

$(f\ o\ h)(x)=\frac{6x^3+44x^2+96x+64}{x^3+12x^2+48x+64}\\ \\(f\ o\ h)(x)=\frac{2(x+2)(3x+4)(x+4)}{(x+4)^2(x+4)}\\ \\(f\ o\ h)(x)=\frac{2(x+2)(3x+4)}{(x+4)^2}\\ \\ \\(g\ o\ f\ o\ h)(x)=3(\frac{2(x+2)(3x+4)}{(x+4)^2})-4\\ \\(g\ o\ f\ o\ h)(x)=\frac{6(x+2)(3x+4)-4(x+4)^2}{(x+4)^2}\\ \\(g\ o\ f\ o\ h)(x)=\frac{6(3x^2+10x+8)-4(x^2+8x+16)}{(x+4)^2}\\ \\(g\ o\ f\ o\ h)(x)=\frac{18x^2+60x+48-4x^2-32x-64}{(x+4)^2}\\ \\(g\ o\ f\ o\ h)(x)=\frac{14x^2+28x-16}{(x+4)^2}$

$(g\ o\ f\ o\ h)(x)=\frac{2(7x^2+14x-8)}{(x+4)^2}$

Soal d

$x=p \\ (h\ o\ g\ o\ f)(p)= -\frac{2}{3}\\ \\-\frac{2(2p^2+p-2)}{p(2p-1)}= -\frac{2}{3}\\ \\ -\frac{2(2p^2+p-2)}{2p^2-p}=-\frac{2}{3}\\ \\-3(2(2p^2+p-2))=-2(2p^2-p)\\ \\ (\frac{-3\times2}{-2})(2p^2+p-2)=2p^2-p \\ \\3(2p^2+p-2)=2p^2-p \\ \\ 6p^2+3p-6=2p^2-p \\ \\ 4p^2+4p-6=0$

$4p^2+4p-6=0...\text{dibagi dengan 2}\\ \\2p^2+2p-3=0\\ \\a=2,\ b=2,\ c=-3\\ \\p_{1,2}= \frac{-b\ ^+_- \sqrt{b^2-4ac}}{2a}\\ \\p_{1,2}= \frac{-2\ ^+_- \sqrt{2^2-4(2)(-3)}}{2(2)}\\ \\p_{1,2}= \frac{-2\ ^+_- \sqrt{4+24}}{2(2)}\\ \\p_{1,2}= \frac{-2\ ^+_- \sqrt{28}}{2(2)}\\ \\p_{1,2}= \frac{-2\ ^+_- \sqrt{4\times7}}{2(2)}\\ \\p_{1,2}= \frac{-2\ ^+_-\ 2\sqrt{7}}{2(2)}\\ \\ \\p_{1}= \frac{-1+ \sqrt{7}}{2}= \frac{1}{2} \sqrt{7}- \frac{1}{2}\\ \\p_{2}= \frac{-1- \sqrt{7}}{2}= - \frac{1}{2}-\frac{1}{2} \sqrt{7}$

Mohon maaf jika ada perhitungan yang salah. Terbuka untuk masukan dan koreksi. Terima kasih.

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