Chapter 12: Q 24. (page 916)

In Exercise, evaluate the given function at the specified points in the domain, and then find the domain and range of the function.
$g (x, y) = \frac{x^{2} + y^{2}}{x + y}, (π, 1), (4, 5)$

Short Answer

Expert verified

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

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

Range is R.

Step by step solution

Step 1. Given information

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

Step 2. Explanation

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

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 $ℝ^{2} - {(x, y) ∣ x + y = 0; x, y \in ℝ}$ while as the range of function is R.

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In Exercise, evaluate the given function at the specified points in the domain, and then find the domain and range of the function.
$g (x, y) = \frac{x^{2} + y^{2}}{x + y}, (π, 1), (4, 5)$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

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In Exercise, evaluate the given function at the specified points in the domain, and then find the domain and range of the function.gx,y=x2+y2x+y,π,1,4,5

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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In Exercise, evaluate the given function at the specified points in the domain, and then find the domain and range of the function.
$g (x, y) = \frac{x^{2} + y^{2}}{x + y}, (π, 1), (4, 5)$