Question

please help me find out what the question marks are

Answer 1

question a

g(x) times $\sqrt{x-5}=\sqrt{x^2-5}$
divide both sides by √(x-5)
$g(x)=\frac{\sqrt{x^2-5}}{\sqrt{x-5}}$
$g(x)=\frac{(\sqrt{x-5})(\sqrt{x^2-5})}{x-5}$
$g(x)=\frac{\sqrt{x^3-5x^2-5x+25}}{x-5}$


question b
$g(x) \space\ times \space\ 1+\frac{1}{x}=x$
divide both sides by 1+1/x
$g(x)=\frac{x}{1+\frac{1}{x}}$
times by x/x
$g(x)=\frac{x^2}{x+1}$


question c
$\frac{1}{x} \space\ times \space\ f(x)=x$
times both sides by x
$f(x)=x^2$


question d
the (f*g)(x) collumn seems to have a domain of only x is greater than or equal to 0?
so maybe we squared the whole thing, ya
wait
ah, ya
basically
$|x|=(\sqrt{x})^2=(\sqrt{x})(\sqrt{x})$
so f(x)=√x


a. $g(x)=\frac{\sqrt{x^3-5x^2-5x+25}}{x-5}$
b. $g(x)=\frac{x^2}{x+1}$
c.$f(x)=x^2$
d. $f(x)=\sqrt{x}$

Answer 2

f   = f(x)
g = g(x)

so.. the first column has the g expression and the second column has the f expression and the third column has their product

a)

$\bf \begin{array}{cccllll} &g(x)&f(x)&(f\cdot g)(x)\\ &------&------&------\\ a)&g&\sqrt{x-5}&\sqrt{x^2-5} \end{array}\implies gf=\sqrt{x^2-5} \\\\\\ g=\cfrac{\sqrt{x^2-5}}{f}}\implies\boxed{g=\cfrac{\sqrt{x^2-5}}{\sqrt{x-5}}}$

b)

$\bf \begin{array}{cccllll} &g(x)&f(x)&(f\cdot g)(x)\\ &------&------&------\\ b)&g&1+\frac{1}{x}&x \end{array}\implies gf=x \\\\\\ g=\cfrac{x}{f}\implies g=\cfrac{x}{1+\frac{1}{x}}\implies g=\cfrac{x}{\frac{x+1}{x}}\implies g=\cfrac{\frac{x}{1}}{\frac{x+1}{x}} \\\\\\ g=\cfrac{x}{1}\cdot \cfrac{x}{x+1}\implies \boxed{g=\cfrac{x^2}{x+1}}$

c)

$\bf \begin{array}{cccllll} &g(x)&f(x)&(f\cdot g)(x)\\ &------&------&------\\ c)&\frac{1}{x}&f&x \end{array}\implies gf=x \\\\\\ f=\cfrac{x}{g}\implies f=\cfrac{x}{\frac{1}{x}}\implies f=\cfrac{\frac{x}{1}}{\frac{1}{x}}\implies f=\cfrac{x}{1}\cdot \cfrac{x}{1}\implies \boxed{f=x^2}$

d)

$\bf \begin{array}{cccllll} &g(x)&f(x)&(f\cdot g)(x)\\ &------&------&------\\ d)&\sqrt{x}&f&|x| \end{array}\implies gf=|x| \\\\\\ f=\cfrac{|x|}{g}\implies f=\cfrac{|x|}{\sqrt{x}}\implies f=\cfrac{\pm x}{\pm\sqrt{x}}\implies \boxed{f=\pm \cfrac{x}{\sqrt{x}}}$


not sure if you are expected to rationalize a) and d), but we could always rationalize the denominator though

a)

$\bf \cfrac{\sqrt{x^2-5}}{\sqrt{x-5}}\cdot \cfrac{\sqrt{x-5}}{\sqrt{x-5}}\implies \cfrac{\sqrt{x^2-5}\cdot \sqrt{x-5}}{(\sqrt{x-5})^2}\implies \cfrac{\sqrt{(x^2-5)(x-5)}}{x-5} \\\\\\ \cfrac{\sqrt{x^3-5x^2-5x+25}}{x-5}$

d)    $\bf \pm\cfrac{x}{\sqrt{x}}\cdot \cfrac{\sqrt{x}}{\sqrt{x}}\implies \pm \cfrac{x\sqrt{x}}{x}$