Solving Functions: Find f+g=5 | [0,inf)

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To solve for functions f and g such that (f+g)(x) = 5 with the domain [0, inf), it is confirmed that (f+g)(x) means f(x) + g(x). Simple solutions include f(x) = 4 and g(x) = 1, or f(x) = 3 and g(x) = 2. More complex examples involve using f(x) = √x and g(x) = 5 - √x, or defining g(x) as 5 - f(x) for any valid f(x) within the specified domain. The discussion emphasizes the flexibility in choosing functions as long as they satisfy the equation and domain requirements. Various function combinations can be explored to meet the criteria.
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Homework Statement


Find two functions f and g such that (f+g)(x) =5 and the domain of f+g is [0,inf)


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The Attempt at a Solution


I am wondering if (f+g)(x) means f(x)+g(x)

If so then can i just say that f(x)=4 and g(x)=1 or f(x)=3 and g(x)=2?
 
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Yes, looks right to me.
 
Thank You :!)
 
Or you could do something exciting like
f(x) = \sqrt{x}, g(x) = 5 - \sqrt{x}
or
f(x) = \sqrt{x} + x^2 - \frac{3 x^{12}}{1 + x}, g(x) = 5 - \sqrt{x} + x^2 - \frac{3 x^{12}}{1 + x}
or any other function f(x) with domain [0, \infty[ and define g(x) = 5 - f(x).
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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