Chapter 14.4B, Problem 1GCE

To determine

To explain:

Thedomain,range symmetry and end behavour of the given function.

Explanation of Solution

Given:

The function is

  $f\left(x\right)=\sqrt{16-x^2}$

Concept used:

The smallest number that the square root of that is still in the real number system, which of course is zero.

Calculation:

Counter example

  $f\left(x\right)=\sqrt{16-x^2}$

Solved the equation

  $\begin{array}{l}16-x^2\ge 0,\\ x^2\le 16\\ so,\\ x\le 4andx\ge -4\\ or\left[-4,4\right]\end{array}$

So, domain

  $D=\left[-4,4\right]$

Draw the graph

  $f\left(x\right)=\sqrt{16-x^2}$

  

Since the graph is even , the graph of $f$ is symmetric with respect to the y-axis . it is similar to the y- axis .firstly graph is increasing the point $\left(-4,0\right)$ to $\left(0,4\right)$ and then decreasing the graph $\left(0,4\right)$ to $\left(4,0\right)$

And the constant point of the graph is $\left(-3.873,1\right)$ to $\left(3.873,1\right)$