what am I even looking at

[livid radish]

so basically I have like 9 questions and I tried doing 1. I'm so screwed.

1. Function f defined as (whatever that is) ≠ -1

1a. find the repeating function for f², f³ and f⁴

[weary ginkgo]

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[solid fog]

that’s like f(f(x)) right

do yk how to do composite functions

ping in reply

[kindred crypt]

okay so f is defined as $f:x\rightarrow\frac{x}{x+1}$
$x\neq 1$ is because if you put it 1 you would divide by 0 and thats not allowed
for a) you need to figure out what $f^2,f^3 and f^4$ are

if you have f to the power of a number (lets say k) it means you need to apply the function k times

So if it's $f^2$ that means you first just do f(x) and the result of that you put into f(x) again

so for $f^2$ first we look at $f(x)=\frac{x}{x+1}$
now the result of that we put into f(x) again, in other words:
$f(\frac{x}{x+1})$
now you need to figure out what happens there, simplify if possible

Apply the same type of thinking for $f^3,f^4$

[glossy sonnet]

entri

[trail magnet]

$f^n(x)=\overbrace{f(f(f(\ldots f(x))))}^{n\hspace{1mm}\text{times}}$\
with this definition we have that:\
$$\boxed{f^2(x)=f\left(\frac{x}{x+1}\right)=\frac{\frac{x}{x+1}}{\frac{x}{x+1}+1}=\frac{x}{x+x+1}=\frac{x}{2x+1}}$$\
$$\boxed{f^3(x)=f(f^2(x))=f\left(\frac{x}{2x+1}\right)=\frac{\frac{x}{2x+1}}{\frac{x}{2x+1}+1}=\frac{x}{x+2x+1}=\frac{x}{3x+1}}$$\
obviously there is a pattern, namely $f^n(x)=\frac{x}{nx+1}$, so finally $$\boxed{f^4(x)=\frac{x}{4x+1}}$$

[glossy sonnet]

vengeance

[solid fog]

?tag helping

[bold bear]

**When helping people, please do not dump the full solution. **
Remember that these people are trying to learn concepts, and our goal should be to reinforce them. Giving them a solution will not help.

**Instead, give them hints and guide them toward the correct solution. **
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[solid fog]

💀 @trail magnet see above

[solid fog]

this is a good example of what to do

[agile fern]

@livid radish

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