Question

Please help me find out what the question marks are

Answer 1

question a

g(x) times
`√(x-5)=√(x^2-5)`
divide both sides by √(x-5)

`g(x)=(√(x^2-5))/(√(x-5))`

`g(x)=((√(x-5))(√(x^2-5)))/(x-5)`

`g(x)=(√(x^3-5x^2-5x+25))/(x-5)`


question b

`g(x) \space\ times \space\ 1+(1)/(x)=x`
divide both sides by 1+1/x

`g(x)=(x)/(1+(1)/(x))`
times by x/x

`g(x)=(x^2)/(x+1)`


question c

`(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|=(√(x))^2=(√(x))(√(x))`
so f(x)=√x


a.
`g(x)=(√(x^3-5x^2-5x+25))/(x-5)`
b.
`g(x)=(x^2)/(x+1)`
c.
`f(x)=x^2`
d.
`f(x)=√(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&√(x-5)&√(x^2-5) \end{array}\implies gf=√(x^2-5) \\\\\\ g=\cfrac{√(x^2-5)}{f}}\implies\boxed{g=\cfrac{√(x^2-5)}{√(x-5)}}`

b)


`\bf \begin{array}{cccllll} &g(x)&f(x)&(f\cdot g)(x)\\ &------&------&------\\ b)&g&1+(1)/(x)&x \end{array}\implies gf=x \\\\\\ g=\cfrac{x}{f}\implies g=\cfrac{x}{1+(1)/(x)}\implies g=\cfrac{x}{(x+1)/(x)}\implies g=\cfrac{(x)/(1)}{(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)&(1)/(x)&f&x \end{array}\implies gf=x \\\\\\ f=\cfrac{x}{g}\implies f=\cfrac{x}{(1)/(x)}\implies f=\cfrac{(x)/(1)}{(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)&√(x)&f&|x| \end{array}\implies gf=|x| \\\\\\ f=\cfrac{g}\implies f=\cfracx{√(x)}\implies f=\cfrac{\pm x}{\pm√(x)}\implies \boxed{f=\pm \cfrac{x}{√(x)}}`


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

a)


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

d)
`\bf \pm\cfrac{x}{√(x)}\cdot \cfrac{√(x)}{√(x)}\implies \pm \cfrac{x√(x)}{x}`