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Can someone please help me with this math equation I have a test tomorrow and need to know how to do this thanks gary.

Graph in question

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Especially for these homework-like problems, people here would like to know what have you tried, and where you are stuck at. For example, for $f(x)$ itself, do you have any ideas on what the domain and range of $f(x)$ might be? – Lord Soth Apr 17 '13 at 21:55

From the graph, you have the points $(0,4)$ and $(3,1)$. So the slope of the line is $\frac{4-1}{0-3}=-1$. Hence, the equation of the line is $f(x)=4-x$. So the domain and range would both be all reals. For $y=\sqrt{f(x)}=\sqrt{4-x}$, we have that $4-x\geq0 \implies x\geq4$. The corresponding range is all non-negative reals (since the square root function must be positive or zero).

The invariant points occur where $y(x)=x.$ For the first one, that means that $4-x=x\implies x=2$. For the second one, that means that $\sqrt{4-x}=x\implies 4-x=x^2\implies x^2+x-4=0$. Using the quadratic formula, you get $x=\frac{-1\pm\sqrt{1^2-4(1)(-4)}}{2(1)}=\frac{-1\pm\sqrt{17}}{2}$, which are not in the domain of f. Hence, there are no invariant points for the second problem.

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and the invariant points? – user73122 Apr 17 '13 at 22:02
what are the points for the 2 graph ? im still confused :( and what I got so far for 2 is that the domain is all real and range is all non negetaie reals 4-x>0 – user73122 Apr 17 '13 at 22:18

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