+118608 hi Isabelle,
It is only a technicality, many people do not use brackes very well.
Here is the question:
Find (f-g)(x) when f(x)=2x+6/3x and g(x)=(sqrt x)-8/3x
By convention this is what is REALLY presented.
\(f(x)=2x+\frac{6}{3}x \quad and \quad g(x)=\sqrt x-\frac{8}{3}x\)
This is how I would have interpreted the meaning
\(f(x)=2x+\frac{6x}{3} \quad and \quad g(x)=\sqrt x-\frac{8x}{3}\)
and this is how Max interpreted it
\(f(x)=\frac{2x+6}{3x} \quad and \quad g(x)=\frac{\sqrt{x} -8}{3x}\)
Do you see now what I meant and why brackets are so important 
+9575 \(\begin{array}{rl}f(x)=\dfrac{2x+6}{3x}&g(x)=\dfrac{\sqrt x - 8}{3x}\\\end{array}\\ (f-g)(x) = f(x) - g(x) = \dfrac{2x - \sqrt x + 14}{3x}\)
.+118608 Hi Max, technically you have answered a different question although it may well be was the asker intended :)
+118608 hi Isabelle,
It is only a technicality, many people do not use brackes very well.
Here is the question:
Find (f-g)(x) when f(x)=2x+6/3x and g(x)=(sqrt x)-8/3x
By convention this is what is REALLY presented.
\(f(x)=2x+\frac{6}{3}x \quad and \quad g(x)=\sqrt x-\frac{8}{3}x\)
This is how I would have interpreted the meaning
\(f(x)=2x+\frac{6x}{3} \quad and \quad g(x)=\sqrt x-\frac{8x}{3}\)
and this is how Max interpreted it
\(f(x)=\frac{2x+6}{3x} \quad and \quad g(x)=\frac{\sqrt{x} -8}{3x}\)
Do you see now what I meant and why brackets are so important
