Q 27. - 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.

$f\left(x,y,z\right)=x^2+y^2+z^2,\left(1,0,-5\right),\left(\frac{1}{2},-1,\frac{1}{3}\right)$

Step-by-Step Solution

Answer

$\begin{array}{rcl}f\left(1,0,-5\right)& =& 26\\ f\left(\frac{1}{2},-1,\frac{1}{3}\right)& =& \frac{49}{36}\end{array}$

Domain is$R^3$ and range is positive real numbers including 0.

Step 1. Given information

Function is $f\left(x,y,z\right)=x^2+y^2+z^2$

Step 2. Explanation

$\begin{array}{rcl}f(1,0,-5)& =& (1{)}^2+(0{)}^2+(-5{)}^2\\ & =& 1+0+25\\ & =& 26\\ f\left(\frac{1}{2},-1,\frac{1}{3}\right)& =& {\left(\frac{1}{2}\right)}^2+(-1{)}^2+{\left(\frac{1}{3}\right)}^2\\ & =& \frac{1}{4}+1+\frac{1}{9}\\ & =& \frac{49}{36}\end{array}$

The given function is a degree 3 polynomial function.

A polynomial function is defined as having a domain and a range that spans the full real numbers without constraint.

As a result, the domain of the given function is $R^3$, whereas the range is positive real numbers including 0.