Given the graph of a function $f(x)$, describe $\sqrt{f(x)}$

Question

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

Answer 1

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.