Q 28. - 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.

$f\left(x,y,z\right)=\frac{x+y+z}{xyz},\left(1,-2,6\right),\left(-3,-2,4\right)$

Step-by-Step Solution

Answer

$\begin{array}{rcl}f\left(1,-2,6\right)& =& -\frac{5}{12}\\ f\left(-3,-2,4\right)& =& -\frac{1}{24}\end{array}$

Domain is $R^3-\left\{\left(0,y,z\right),\left(x,0,z\right),\left(x,y,0\right)\right\}$ and range is R.

Step 1. Given information

Function is $f\left(x,y,z\right)=\frac{x+y+z}{xyz}$

Step 2. Explanation

$\begin{array}{rcl}f(1,-2,6)& =& \frac{\left(1\right)+(-2)+\left(6\right)}{\left(1\right)(-2)\left(6\right)}\\ & =& \frac{5}{-12}\\ & =& -\frac{5}{12}\\ f(-3,-2,4)& =& \frac{(-3)+(-2)+\left(4\right)}{(-3)(-2)\left(4\right)}\\ & =& \frac{-1}{24}\\ & =& -\frac{1}{24}\end{array}$

For domain:

width="52" style="max-width: none; vertical-align: -4px;" $xyz=0$

x= 0 or y = 0 or z = 0

These points are to be excluded from function's domain.

So, domain is $R^3-\left\{\left(0,y,z\right),\left(x,0,z\right),\left(x,y,0\right)\right\}$ range is R.