Finding explicit forms of a function

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

I have the following answers for each part and I am wondering whether each is please correct:
(a) $\frac{x}{x + 1} + \frac{1}{x - 2}, x \neq -1, 2$
(b) $\frac{x}{(x + 1)(x - 2)}, x \neq -1, 2$
(c) $\frac{2x}{x + 1}, x \neq -1$
(d) $\frac{-x}{1 - x}, x \neq 1$
(e) $\frac{\frac{1}{x - 2}}{\frac{1}{x - 2} + 1}, x \neq 2$
(g) $\frac{\frac{x}{x + 1}}{\frac{1}{x - 2}}, x \neq -1, 2$
(h) $\frac{x^2}{(x + 1)^2}, x \neq - 1$
(i) $\frac{\frac{x}{x + 1}}{\frac{x}{x + 1} + 1}, x \neq -1$

Thanks!

 

[Hill]

I think that:
(e) the domain is incorrect,
(f) missing,
(i) the domain is incorrect.

 

[docnet]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 346376
I have the following answers for each part and I am wondering whether each is please correct:

(e) $\frac{\frac{1}{x - 2}}{\frac{1}{x - 2} + 1}, x \neq 2$
(g) $\frac{\frac{x}{x + 1}}{\frac{1}{x - 2}}, x \neq -1, 2$
(h) $\frac{x^2}{(x + 1)^2}, x \neq - 1$
(i) $\frac{\frac{x}{x + 1}}{\frac{x}{x + 1} + 1}, x \neq -1$

Thanks!

(e)
\begin{align*}\frac{\frac{1}{x - 2}}{\frac{1}{x - 2} + 1}=&\frac{\frac{1}{x - 2}}{\frac{x-1}{x - 2} }\\
=& \frac{1}{x - 2}\cdot \frac{x - 2}{x-1}\\
=&\frac{1}{x-1}.\end{align*}
And so $x\neq 1$.

(g) \begin{align*}\frac{\frac{x}{x + 1}}{\frac{1}{x - 2}}=& \frac{x}{x + 1}\cdot (x-2).
\end{align*}

And $x\neq -1$.

(i) This one is not so different from (e): simplify the fraction first.

 

[SammyS]

(e)
\begin{align*}\frac{\frac{1}{x - 2}}{\frac{1}{x - 2} + 1}=&\frac{\frac{1}{x - 2}}{\frac{x-1}{x - 2} }\\
=& \frac{1}{x - 2}\cdot \frac{x - 2}{x-1}\\
=&\frac{1}{x-1}.\end{align*}
And so $x\neq 1$.

For (e):
The Domain of $\displaystyle f \circ g \$ contains neither $2$ nor $1$ .

 

[docnet]

For (e):
The Domain of $\displaystyle f \circ g \$ contains neither $2$ nor $1$ .

You're right, you want the domain of the composite function before simplification. I guess they're not the same functions because their domains are different, although they can be algebraically manipulated to be equivalent. So my post contains that mistake. .

 

[SammyS]

You're right, you want the domain of the composite function before simplification. I guess they're not the same functions because their domains are different, although they can be algebraically manipulated to be equivalent. So my post contains that mistake. .

Similarly, your answer for part (g) is incorrect.

Also, it would be better if we could get OP to do these corrections, rather than see him/her merely react with "Like", etc.

 

[SammyS]

I think that:
(e) the domain is incorrect,
(f) missing,
(g) the domain is incorrect,
(i) the domain is incorrect.

The domain given for (g) was correct in the OP,

Come on, @ChiralSuperfields ! Defend your solutions. Sometimes they're correct.

 

[Hill]

The domain given for (g) was correct in the OP,

Come on, @ChiralSuperfields ! Defend your solutions. Sometimes they're correct.

Thank you. I have corrected my reply in the post #2.