Chapter 12: Q 26. (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.
$f (x, y) = \frac{\sin (x - y)}{x - y}, (0, π), (\frac{π}{2}, \frac{π}{3})$

Short Answer

Expert verified

$f (0, π) = 0 f (\frac{π}{2}, \frac{π}{3}) = \frac{3}{π}$

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

Range is R.

Step by step solution

Step 1. Given information

Function is $f (x, y) = \frac{\sin (x - y)}{x - y}$

Step 2. Explanation

$\begin{array}{rcl} f (0, π) & = & \frac{\sin (0 - π)}{0 - π} \\ = & \frac{- \sin (π)}{- π} \\ = & 0 \\ f (\frac{π}{2}, \frac{π}{3}) & = & \frac{\sin (\frac{π}{2} - \frac{π}{3})}{(\frac{π}{2} - \frac{π}{3})} \\ = & \frac{\frac{1}{2}}{\frac{π}{6}} \\ = & \frac{3}{π} \end{array}$

For domain, denominator can not be equal to 0.

$\begin{array}{rcl} x - y & = & 0 \\ x & = & y \end{array}$

So, domain is $R^{2} - \{(x, y) | x - y = 0\}$

Let $w = x - y$

So, width="93">F(w)=sinww

$\begin{array}{rcl} F^{'} (w) & = & \frac{w \cos w - \sin w}{w} \\ 0 & = & \frac{w \cos w - \sin w}{w} \\ w \cos w - \sin w & = & 0 \\ w & = & \tan w \end{array}$

So, one of the solution is 0.

Function F(w) is not defined for this solution.

Also,

$\lim_{w \to 0} \frac{\sin w}{w} = 1$

So, maximum value of F(w) is 1.

Take w = c

width="58">c=tanc

$\begin{array}{rcl} F (c) & = & \frac{\sin c}{c} \\ = & \cos c \end{array}$

So, minimum value of function is $\cos c$

Using the maximum and minimum value of function, the range of function is written as [c, 1] where c is the non-zero solution of width="66">w=tanw

So, range of the 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.
$f (x, y) = \frac{\sin (x - y)}{x - y}, (0, π), (\frac{π}{2}, \frac{π}{3})$

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.fx,y=sinx-yx-y,0,π,π2,π3

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.
$f (x, y) = \frac{\sin (x - y)}{x - y}, (0, π), (\frac{π}{2}, \frac{π}{3})$