Q 24. - Calculus

Question

In Exercise, evaluate the given function at the specified points in the domain, and then find the domain and range of the function.

$g\left(x,y\right)=\frac{x^2+y^2}{x+y},\left(\mathrm{\pi },1\right),\left(4,5\right)$

Step-by-Step Solution

Answer

$\begin{array}{rcl}f(\pi ,1)& =& \frac{{\pi }^2+1}{\pi +1}\\ f(4,5)& =& \frac{41}{9}\end{array}$

Domain is $R^2-\left\{\left(x,y\right)|x+y=0:x,y\in R\right\}$.

Range is R.

Step 1. Given information

Function is $g\left(x,y\right)=\frac{x^2+y^2}{x+y}$

Step 2. Explanation

$\begin{array}{rcl}f(\pi ,1)& =& \frac{(\pi {)}^2+(1{)}^2}{\pi +1}\\ & =& \frac{{\pi }^2+1}{\pi +1}\\ f(4,5)& =& \frac{(4{)}^2+(5{)}^2}{4+5}\\ & =& \frac{16+25}{9}\\ & =& \frac{41}{9}\end{array}$

For domain:

$\begin{array}{rcl}x+y& =& 0\\ x& =& -y\end{array}$

As a result, the points on the line x + y = 0 must be eliminated from the function's domain. There are no limitations on the output values.

So, domain of the given function is ${\mathrm{ℝ}}^2-\left\{\right(x,y)\mid x+y=0;x,y\in \mathrm{ℝ}\}$ while as the range of function is R.