91Ó°ÊÓ

Multivariable functions are functions that depend on more than one variable. For example, in the given exercise:
- The function f(x, y) = x - y depends on two variables: x and y.
Such functions are often visualized as surfaces or hypersurfaces in higher dimensions.

Understanding how to work with multivariable functions will greatly help in solving more complex calculus problems. In the context of our exercise, correctly identifying and evaluating these functions is a crucial step towards finding solutions.

Function composition is combining two functions where the output of one function becomes the input of the other.
For example, if you have two functions, g(t) and h(s), the composition (g â—¦ h)(s) means that you first apply h to the input s and then apply g to the result of h(s).
In our exercise, one instance of function composition is the expression \( (g \circ f)(5,1) \). This means you first apply f to (5, 1), and then apply g to the result obtained from f.
Here is the step-by-step breakdown:
- Compute f(5, 1): f(5, 1) = 5 - 1 = 4
- Then apply g to this result: g(4) = √4 = 2

So, (g â—¦ f)(5,1) = 2.
Remember, understanding function composition requires careful tracking of each intermediate step.

Approaching calculus problems with step-by-step solutions is a great way to understand each part of the process. Breaking down the problem ensures that you catch and understand every detail.
Here's how to solve each part of our given exercise systematically:
- Identify the given functions.
- Carefully follow each step to compute intermediary results.
For example, in part (b), we computed f(h(3), g(9)). The steps are:
- Compute h(3): h(3) = 3^2 = 9
- Compute g(9): g(9) = √9 = 3
- Substitute into f: f(9, 3) = 9 - 3 = 6
Thus, f(h(3), g(9)) = 6.
Using a systematic, step-by-step approach ensures no steps are skipped and that the final result is accurate.

Calculus problems often involve different types of functions such as polynomials, exponentials, and trigonometric functions. In our exercise, we deal with functions involving square roots and squares.
These problems require you to understand various operations and function behaviors. For instance:
- When dealing with the composition of functions, particularly when nested, it's essential to handle each layer one at a time.
- Applying the chain rule or other calculus techniques might be necessary for more advanced problems.
To illustrate, let’s take part (d) from our problem: g((h ◦ f)(x, y)).
- First, compute f(x, y): f(x, y) = x - y
- Then apply h: h(f(x, y)) = (x - y)^2
- Finally apply g: g((x - y)^2) = √((x - y)^2) = |x - y|
The absolute value arises because the square root of a square can return either the positive or negative root value.
By practicing different combinations and compositions of functions, you become more adept at handling various calculus problems.