Pick any exercise in Chapter 12 of the calculus textbook. Post the solution or your attempt. Discuss any issues you might have had. What were the most valuable elements you took away from this course?
I chose Problems 7.1.9-14:
Exercises 7-14, give the domain and range of the multivariable function.
\(\boldsymbol{9. f(x,y) = x - 2y}\)
The domain is all real numbers for both \(x\) and \(y\).
The range is also all real numbers. You can come up with values for \(x\) and \(y\) that could give you any possible result.
\(\boldsymbol{10. f(x,y) = \frac{1}{x + 2y}}\)
The domain is any combination of \(x\) and \(y\) except any combination where \(x + 2y \neq 0\).
The range is all real values except \(0\).
\(\boldsymbol{11. f(x,y) = \frac{1}{x^2 + y^2 + 1}}\)
The domain is all real numbers for both \(x\) and \(y\).
The range is greater than \(0\) and less than or equal to \(1\). The lowest value possible for the denominator is 1, and the highest value is infinity.
\(\boldsymbol{12. f(x,y) = \sin x \cos y}\)
The domain is all real numbers for both \(x\) and \(y\).
The range for \(\sin x\) is \(-1\) to \(1\), and range for \(\cos y\) is also \(-1\) to \(1\), therefore the range for the product of the two is also \(-1\) to \(1\),
\(\boldsymbol{13. f(x,y) = \sqrt{9 - x^2 - y^2}}\)
The domain is any \(x\) and \(y\) where \(x^2 + y^2 \leq 9\), because the denominator cannot be zero and you can’t take the square root of a negative.
The range is \(\geq 0\) and \(\leq 3\); if \(x = 0\) and \(y = 0\) then the max value is \(3\), and any other values would be between \(0\) and \(3\).
\(\boldsymbol{14. f(x,y) = \frac{1}{\sqrt{x^2 + y^2 - 9}}}\)
The domain is any \(x\) and \(y\) where \(x^2 + y^2 \geq 9\) and \(\sqrt{x^2 + y^2 - 9} \neq 0\).
The range is all real numbers greater than \(0\).