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Question Number 81598 by zainal tanjung last updated on 03/Mar/20

fog(x)=8x+3 g(x)=2x−1 f(x)=.....? gof(x)=6x+1 g(x)=5x+1 f(x)=.....? gof(x)=7x+9 f(x)=5x+2 g(x)=.....? f(x)=3x−8 gof(x)=8x+3 g(x)=.....? gof(x)=5x+1 f(x)=3x g(x)=....?

$$\mathrm{fog}\left(\mathrm{x}\right)=\mathrm{8x}+\mathrm{3} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{2x}−\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=.....? \\ $$$$ \\ $$$$\mathrm{gof}\left(\mathrm{x}\right)=\mathrm{6x}+\mathrm{1} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{5x}+\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=.....? \\ $$$$ \\ $$$$\mathrm{gof}\left(\mathrm{x}\right)=\mathrm{7x}+\mathrm{9} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{5x}+\mathrm{2} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=.....? \\ $$$$ \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3x}−\mathrm{8} \\ $$$$\mathrm{gof}\left(\mathrm{x}\right)=\mathrm{8x}+\mathrm{3} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=.....? \\ $$$$ \\ $$$$\mathrm{gof}\left(\mathrm{x}\right)=\mathrm{5x}+\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3x} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=....? \\ $$$$ \\ $$

Commented by Tony Lin last updated on 14/Feb/20

let f(x)=ax+b a(2x−1)+b=8x+3 a=4 , b=7 ⇒f(x)=4x+7

$${let}\:{f}\left({x}\right)={ax}+{b} \\ $$$${a}\left(\mathrm{2}{x}−\mathrm{1}\right)+{b}=\mathrm{8}{x}+\mathrm{3} \\ $$$${a}=\mathrm{4}\:,\:{b}=\mathrm{7} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{4}{x}+\mathrm{7} \\ $$

Commented by mr W last updated on 14/Feb/20

fog(x)=8x+3=4(2x−1)+7 f(2x−1)=4(2x−1)+7 f(x)=4x+7

$${fog}\left({x}\right)=\mathrm{8}{x}+\mathrm{3}=\mathrm{4}\left(\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{7} \\ $$$${f}\left(\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{4}\left(\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{7} \\ $$$${f}\left({x}\right)=\mathrm{4}{x}+\mathrm{7} \\ $$

Commented by Zainal Arifin last updated on 14/Feb/20

You are Right!

$$\mathrm{You}\:\mathrm{are}\:\mathrm{Right}! \\ $$

Commented by mathmax by abdo last updated on 14/Feb/20

let h(x)=8x+x we have fog(x)=h(x)⇒f(x)=hog^(−1) (x) we have g(x)=2x−1 =y ⇒x=g^(−1) (y) and 2x−1=y⇒ x=((1+y)/2) ⇒g^(−1) (x)=((1+x)/2) ⇒f(x)=h(((x+1)/2))=8(((x+1)/2))+3 =4x+4+3 =4x+7 ⇒★f(x)=4x+7★

$${let}\:{h}\left({x}\right)=\mathrm{8}{x}+{x}\:\:{we}\:{have}\:{fog}\left({x}\right)={h}\left({x}\right)\Rightarrow{f}\left({x}\right)={hog}^{−\mathrm{1}} \left({x}\right) \\ $$$${we}\:{have}\:{g}\left({x}\right)=\mathrm{2}{x}−\mathrm{1}\:={y}\:\Rightarrow{x}={g}^{−\mathrm{1}} \left({y}\right)\:{and}\:\mathrm{2}{x}−\mathrm{1}={y}\Rightarrow \\ $$$${x}=\frac{\mathrm{1}+{y}}{\mathrm{2}}\:\Rightarrow{g}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}+{x}}{\mathrm{2}}\:\Rightarrow{f}\left({x}\right)={h}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)=\mathrm{8}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)+\mathrm{3} \\ $$$$=\mathrm{4}{x}+\mathrm{4}+\mathrm{3}\:=\mathrm{4}{x}+\mathrm{7}\:\Rightarrow\bigstar{f}\left({x}\right)=\mathrm{4}{x}+\mathrm{7}\bigstar \\ $$

Commented by mathmax by abdo last updated on 14/Feb/20

h(x)=8x+3

$${h}\left({x}\right)=\mathrm{8}{x}+\mathrm{3} \\ $$

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