use the following information to answer written - response question 1. the graph of y = f(x) is shown below…

Explanation:

Step1: Understand invariant points

Invariant points of $y = \sqrt{f(x)}$ with respect to $y = f(x)$ are points where $f(x)=\sqrt{f(x)}$. Let $t = f(x)$, then $t=\sqrt{t}$, which gives $t^2 - t=0$, so $t(t - 1)=0$. Thus $t = 0$ or $t = 1$.

Step2: Find $x$ - values for $f(x)=0$

From the graph of $y = f(x)$, we find the $x$ - values where $y = 0$. The graph of $y = f(x)$ intersects the $x$ - axis at $x=-2$ and $x = 2$.

Step3: Find $x$ - values for $f(x)=1$

From the graph of $y = f(x)$, we find the $x$ - values where $y = 1$. The graph of $y = f(x)$ intersects the line $y = 1$ at $x=-4$ and $x=4$.

Step4: Sketch $y=\sqrt{f(x)}$

For $y = f(x)\geq0$, when $0\leq f(x)\leq1$, $\sqrt{f(x)}\geq f(x)$. When $f(x)>1$, $\sqrt{f(x)}<f(x)$. The graph of $y=\sqrt{f(x)}$ will have the same $x$ - intercepts as $y = f(x)$ (since $\sqrt{0}=0$) and the points where $f(x) = 1$ will be invariant. Also, the parts of the graph of $y = f(x)$ below the $x$ - axis will not be part of the graph of $y=\sqrt{f(x)}$ because the square - root of a negative number is not real.

Answer:

The invariant points are $(-4,1),(-2,0),(2,0),(4,1)$.