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Find $g$

a) If $g∘(2f)=f+h$ where $f(x)=2x + 5$ and $h(x)=x^3 -2x$

b) If $(2f)∘g=f+h$ where $f(x)=\ln(x+2)$ and $h(x)=\sin(x^2)$

Thanks

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1 Answer 1

(a) Write out $f+h$ (that you can do directly). Then write out $2f$. Then try to expresss $f+h$ in terms of $2f$. Your expression is what $g$ does.

For example, if we have $f(x) = 3x+1$ and $h(x) = 36x^2+21x+3$, then $f+h = 36x^2 + 24x+4$. On the other hand, $2f = 6x+2$. Can we write $f+h$ in terms of $2f$? Yes: $(6x+2)^2 = 36x^2 + 24x + 4$, so $g(u) = u^2$ would satisfy $g\circ(2f)=f+h$.

(b) Similar to (a), but now you are doing $(2f)\circ g$ with the given $f$ and $h$.

In short: Write it out, figure it out.

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i'm sorry but i still can't figure out you gave a rather silly example –  user7143 Feb 23 '11 at 19:28
1  
@user7143: I'm sorry, but you have to show you are doing some of the work yourself. Show what you've done by editing the question, and where you are stuck, and then people can nudge you in the right direction rather than work out your homework for you. –  Arturo Magidin Feb 23 '11 at 19:41
    
in a) i'm stuck here $(4x+10)^3=x^3+5$ that wasn't very helpful –  user7143 Feb 23 '11 at 19:58
    
@user7143: Sorry, but I don't understand how you got to that. Again: please edit your question by writing out everything you've done so far up to the point you are stuck. Editing your question will give you more space to explain what is going on, so you don't have to rely on the shortened space for comments, and it will let you provide the entire line of thinking you have been following, rather than just the punchline (which, having not heard the rest of the joke, I don't really understand). –  Arturo Magidin Feb 23 '11 at 20:13
    
@user7143: The reason I wrote all of that is that if I were trying to do this problem, I don't believe I would have every gotten to an equation of the form $(4x+10)^3 = x^3+5$, which is why I don't know how you got there. Hard to get someone unstuck if you have no inkling how they got stuck in the first place. –  Arturo Magidin Feb 23 '11 at 23:12

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