Q 25. - 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.

$g\left(x,y\right)=\frac{\ln \left(xy\right)}{\sqrt{1-x}},\left(-e,-1\right),\left(\frac{1}{2},2\right)$

Step-by-Step Solution

Answer

Domain is $\left\{\right(x,y)\mid x<0;y<0\}\cup \left\{\right(x,y)\mid 0<x<1;y>0\}$

The range is R.

$\begin{array}{rcl}g(-e,-1)& =& \frac{1}{\sqrt{1+e}}\\ g(-e,-1)& =& 0\end{array}$

Step 1. Given information

Function is $g\left(x,y\right)=\frac{\ln \left(xy\right)}{\sqrt{1-x}}$

Step 2. Explanation

$\begin{array}{rcl}g(-e,-1)& =& \frac{\ln \left(\right(-e\left)\right(-1\left)\right)}{\sqrt{1-(-e)}}\\ & =& \frac{\ln \left(e\right)}{\sqrt{1+e}}\\ & =& \frac{1}{\sqrt{1+e}}\\ g(-e,-1)& =& \frac{\ln \left(\left(\frac{1}{2}\right)\left(2\right)\right)}{\sqrt{1-\left(\frac{1}{2}\right)}}\\ & =& \frac{\ln \left(1\right)}{\sqrt{\frac{1}{2}}}\\ & =& 0\end{array}$

Another goal is to figure out the function's domain and range.

A logarithmic function is used in the function. The positive real numbers are the domain of a logarithmic function. Make the logarithmic function's parameter greater than 0.

$x y >0$

$x, y>0$ or $x,y<0$

$\begin{array}{rcl}1-x& >& 0\\ x& <& 1\end{array}$

Domain is $\left\{\right(x,y)\mid x<0;y<0\}\cup \left\{\right(x,y)\mid 0<x<1;y>0\}$

Range is R as there is no constraint over the output values.