Chapter 1.4, Problem 63E

To determine

The domain of the function.

Answer to Problem 63E

The domain of the function will be all real values of $x$ where $x>2$ .

Explanation of Solution

Given:

The right angled triangle in first quadrant with base $a$ and height b.

  

Formula used:

The area of right angled triangle, A is $\frac{1}{2}\times \text{base}\times \text{height}$

Calculation:

Consider following right angled triangle in first quadrant with base $a$ and height b.

  

Aline is passing through coordinate (2, 1).

A right angled triangle with base x and height y as there are infinite lines passing through point (2, 1).

The area of right angled triangle, A is $\frac{1}{2}\times \text{base}\times \text{height}$

  $A=\frac{1}{2}xy$

Now determining the functionof A in terms of X.

Making to get y in terms ofX

The slope of a line is $\frac{y_2-y_1}{x_2-x_1}$

Slope of line with coordinates $\left(0,y\right)\text{and}\left(2,1\right)\text{is}\frac{y-1}{0-2}$

Slope of line with coordinates $\left(x,0\right)\text{and}\left(2,1\right)\text{is}\frac{1-0}{2-x}$

  $\begin{array}{l}\frac{y-1}{0-2}=\frac{1-0}{2-x}\\ y-1=\frac{-2}{2-x}\\ y=\frac{2}{x-2}+1\\ y=\frac{2+x-2}{x-2}\\ y=\frac{x}{x-2}\end{array}$

Thus, we get

  $\begin{array}{l}A=\frac{1}{2}xy\\ A=\frac{1}{2}x\left(\frac{x}{x-2}\right)\\ A=\frac{x^2}{2\left(x-2\right)}\end{array}$

This is the required answer.

Now, finding the domain of above function as,

As the right angled triangle is in first quadrant, then X can only takes positive values, that is

  $x>0$

It includes only those values in the domain where denominator is non-zero.

  $\begin{array}{l}x-2=0\\ x=2\end{array}$

For x=2, the denominator becomes zero.

So, exclude x=2

Hence, the domain of this function is all real values of $x$ where $x>2$ .

Conclusion:

The domain of the function will be all real values of $x$ where $x>2$ .