Question

Let \(g(x,y) = \cos (x + 2y)\)

1. Evaluate \(g(2, - 1)\)
2. Find the domain of g
3. Find the e range of g

Short answer

a) The g(2,-1) = 1

b) The domain function of g(x,y) is entire plane (R­­­­²)

c) The range of g(-1,1)

Step-by-step solution

Step 1:- Consider the function

Given \(g(x,y) = \cos (x + 2y)\)

Step 2:- Substitution

Let x=2 and y=-1 in the function g(x,y)

\(\begin{aligned}{l} \Rightarrow g(2,1) &= \cos (2 + 2( - 1))\\ &= \cos (2 - 2)\\ &= \cos (0)\\ \Rightarrow g(2, - 1) &= 1\end{aligned}\)

Therefore,

b) The domain function of g(x,y) is entire plane of R².

Step 1:- Consider the function

Given function \(g(x,y) = \cos (x + 2y)\)

Step 2:- Consider \(x + 2y = z\)

The \(\cos (x + 2y)\) is a function with rel value of the cos function.

The \(x + 2y = z\) which is a real number or belong to real part and there is no imaginary part.

The linear function x+2y is defined for real valoue of x,y.

Hence, the domain of the function g(x,y) the entire palne R²

c)The range of the function g(x,y)(-1,1)

Step 1:- Determine the range of \((x + 2y)\)

Consider \((x + 2y)\)

x belongs to R, where R is real

y belongs to R

\( \Rightarrow (x + 2y)\)belongs to R

Therefore, the range of \((x + 2y)\) is R.

Step 2:- Determining the range of cos (x+2y)

Since, the range of \((x + 2y)\) is real then the cos function of \((x + 2y)\) lies between -1 to +1.

Therefore, the range of g is \(( - 1,1)\)