Problem 30 - The Calculus with Analytic Geometry

Question

Given \(f(x, y)=x / y^{2}, g(x)=x^{2}, h(x)=\sqrt{x}\). Find (a) \((h \circ f)(2,1) ;\) (b) \(f(g(2), h(4)) ;\) (c) \(f\left(g(\sqrt{x}), h\left(x^{2}\right)\right) ;\) (d) \(h((g \circ f)(x, y)) ;\) (e) \((h \circ g)(f(x, y))\).

Step-by-Step Solution

Answer

(a) \( \sqrt{2} \); (b) \( 1 \); (c) \( \frac{1}{x} \); (d) \( \frac{x}{y^2} \); (e) \( \frac{x}{y^2} \)

Step 1: Understanding the functions

Identify the functions given: \( f(x, y) = \frac{x}{y^2} \), \( g(x) = x^2 \), and \( h(x) = \sqrt{x} \).

Step 1(a) - Calculate \(h \circ f(2, 1)\)

Firstly, evaluate \(f(2, 1)\). \( f(2, 1) = \frac{2}{1^2} = 2 \). Now evaluate \(h(2)\). \( h(2) = \sqrt{2} \). So, \((h \circ f)(2,1) = h(f(2, 1)) = h(2) = \sqrt{2}\).

Step 1(b) - Calculate f(g(2), h(4))

First, find \(g(2)\) and \(h(4)\). \( g(2) = 2^2 = 4 \). \( h(4) = \sqrt{4} = 2 \). Now substitute these into \(f\). \( f(g(2), h(4)) = f(4, 2) = \frac{4}{2^2} = \frac{4}{4} = 1 \).

Step 1(c) - Calculate f(g(\sqrt{x}), h(x^2))

First, find \(g(\sqrt{x})\) and \(h(x^2)\). \( g(\sqrt{x}) = (\sqrt{x})^2 = x \). \( h(x^2) = \sqrt{x^2} = x \). Now substitute these into \(f\). \( f(g(\sqrt{x}), h(x^2)) = f(x, x) = \frac{x}{x^2} = \frac{1}{x} \).

Step 1(d) - Calculate h((g \circ f)(x, y))

First, find \(g(f(x, y))\). \( f(x, y) = \frac{x}{y^2} \). Now find \(g(\frac{x}{y^2})\). \( g(\frac{x}{y^2}) = (\frac{x}{y^2})^2 = \frac{x^2}{y^4} \). Now substitute this value into \(h\). \( h\left( \frac{x^2}{y^4} \right) = \sqrt{\frac{x^2}{y^4}} = \frac{x}{y^2} \).

Step 1(e) - Calculate (h \circ g)(f(x, y))

First, evaluate \(f(x,y)\). \( f(x, y) = \frac{x}{y^2} \). Now find \(g(\frac{x}{y^2})\). \( g(\frac{x}{y^2}) = (\frac{x}{y^2})^2 = \frac{x^2}{y^4} \). Evaluate \(h(\frac{x^2}{y^4})\). \( h\left( \frac{x^2}{y^4} \right) = \sqrt{\frac{x^2}{y^4}} = \frac{x}{y^2} \). Thus, \((h \circ g)(f(x, y)) = h\left(g\left(\frac{x}{y^2}\right)\right) = h\left(\frac{x^2}{y^4}\right) = \frac{x}{y^2} \).

Key Concepts

Function Composition

Function composition is when you apply one function to the results of another function. Imagine you have two functions, say, \( f \) and \( g \). The composition of \( g \) with \( f \), denoted as \( (g \circ f)(x) \), means that you first apply \( f \) to \( x \) and then apply \( g \) to the result of \( f \).
Using an example from our exercise, when finding \( (h \circ f)(2, 1) \), we first evaluate \( f(2, 1) \) which gives us 2. Then, we use that result as the input for function \( h \), resulting in \( h(2) = \sqrt{2} \).
By chaining functions together, you can simplify complex problems step-by-step.
This is particularly useful when dealing with functions that manipulate different variables or multiple stages of transformations.

Multivariable Functions

Multivariable functions are functions with more than one input. For instance, \( f(x, y) = \frac{x}{y^2} \) takes both \( x \) and \( y \) as inputs.
When working with these functions in composition, it’s important to carefully manage how each input variable is transformed.
Taking another example from the exercise, for \( f(g(2), h(4)) \), we first find \( g(2) \) which equals 4, and \( h(4) \) which equals 2. These results are then used in \( f \) as the inputs, making it \( f(4, 2) = \frac{4}{2^2} = 1 \).
These step-by-step evaluations help break down complicated expressions into manageable parts, making them easier to solve and understand.

Calculus Operations

Calculus operations involve differentiation and integration, but they often start with understanding basic function manipulation.
The exercise leverages basic functions but these principles extend to more complex calculus scenarios.
For instance, when working with expressions like \( f(g(\sqrt{x}), h(x^2)) \), you first simplify each inner function before dealing with the outer function.
From the example, you simplify to \( g(\sqrt{x}) = x \) and \( h(x^2) = x \), then they become inputs of \( f \). The final result is \( \frac{1}{x} \).
This structured approach ensures clarity before progressing to more advanced calculus operations like finding derivatives or integrals of composed functions.