Chapter 1.1, Problem 50E

To determine

To find: An expression for the function whose graph is the top half of the circle $x^2+{\left(y-2\right)}^2=4$.

Answer to Problem 50E

The function of the top half of the circle $x^2+{\left(y-2\right)}^2=4$ is $\underset{\_}{f\left(x\right)=2+\sqrt{4-x^2}}$ where $-2\le x\le 2$.

Explanation of Solution

Given:

The graph is the top half of the circle $x^2+{\left(y-2\right)}^2=4$.

Calculation:

Solve the equation of the circle $x^2+{\left(y-2\right)}^2=4$ and solve for y.

$\begin{array}{l}{x}^2+{\left(y-2\right)}^2=4\\ {\left(y-2\right)}^2=4-x^2\\ y-2=\pm \sqrt{4-x^2}\\ y=2\pm \sqrt{4-x^2}\end{array}$

Since the graph is the top half of the circle, consider the root with a positive radical. Therefore, the function of the top half of the circle is $f\left(x\right)=2+\sqrt{4-x^2}$.

The expression inside the square root cannot be negative. This implies,

$\begin{array}{l}4-x^2\ge 0\\ 4\ge x^2\\ \pm 2\ge x\end{array}$

Thus, the domain is $-2\le x\le 2$.

Therefore, the function of the top half of the circle $x^2+{\left(y-2\right)}^2=4$ is $\underset{\_}{f\left(x\right)=2+\sqrt{4-x^2}}$ where $-2\le x\le 2$.