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## anonymous one year ago Find gof if f(x)=x+(1/x) and g(x)= (x+1)/(x+2)

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1. princeharryyy

g(f(x))= (f(x)+1)/(f(x)+2)

2. princeharryyy

just out the value of f(x) in place of f(x) in above equation n solve

3. princeharryyy

@Theanitrix

4. princeharryyy

put*

5. anonymous

I know that, I just am not sure how to solve it

6. princeharryyy

ok

7. anonymous

the fraction confuses me

8. anonymous

$(g \circ f) = g\left(x+\frac{1}{x}\right) = \frac{\left(x+\dfrac{1}{x}\right)+1}{\left(x+\dfrac{1}{x}\right)+2}$

9. princeharryyy

well, you got the answer now, I guess.

10. anonymous

That isn't the answer.

11. princeharryyy

that's almost the final answer @Jhannybean

12. anonymous

the answer would be (x^2+x+1)/(x+1)^2

13. anonymous

but i want to know how they got that answer

14. anonymous

multiply both the numerator and denominator by $$x$$. $\frac{\left(x+\dfrac{1}{x}\right)+1}{\left(x+\dfrac{1}{x}\right)+2} \cdot \frac{x}{x}$

15. princeharryyy

(g∘f)=g(x+1/x)=(x^2+1+x)/(x^2+2+x)

16. anonymous

is that because it is the denominator of 1/x? @Jhannybean

17. anonymous

Yes @Theanitrix

18. anonymous

okay so would that same thing apply to fof?

19. anonymous

$(f\circ f)= f\left(x+\frac{1}{x}\right) = \left(x+\frac{1}{x}\right)+\dfrac{1}{x+\dfrac{1}{x}}$

20. anonymous

but when you're trying to solve this... set all that complicated junk = $$a$$ or any other variable. find the common denominator, and simplify.

21. anonymous

$\text{let a}=x+\frac{1}{x}$$(f\circ f) = a+\frac{1}{a} = \frac{a^2+1}{a} = \dfrac{\left(x+\dfrac{1}{x}\right)^2+1}{\left(x+\dfrac{1}{x}\right)}$

22. anonymous

$=\frac{x^2+\dfrac{1}{x^2}+2 +1}{\dfrac{x^2+1}{x}} =\frac{x^2+\dfrac{1}{x^2}+3}{\dfrac{x^2+1}{x}}$ So now you'd want to multiply both numerator and denominator by $$\dfrac{x}{x}$$

23. anonymous

i did it a different way

24. anonymous

and got the answer x^4+3x^2+1/x(x^2+1)

25. anonymous

Thats correct

26. anonymous

okay, i understood your approach as well so thank you

27. anonymous

$\frac{x^2+\dfrac{1}{x^2}+3}{\dfrac{x^2+1}{x}} \cdot \frac{x}{x} =\frac{\dfrac{x^4+3x^2+1}{x}}{x^2+1} = \color{red}{\frac{x^4+3x^2+1}{x(x^2+1)}}$

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