Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

To nd \(f(g(x))\), apply \(g\) to \(x\) and then use the output of \(g\) as the input of \(f .\) Work from the inside out. Let \(f(x)=x^{2}, g(x)=1 / x\), and \(h(x)=3 x+1\) Worked example: $$f(g(h(x)))=f(g(3 x+1))=f\left(\frac{1}{3 x+1}\right)=\left(\frac{1}{3 x+1}\right)^{2}$$ Find the following. (a) \(f(g(x))\) (b) \(g(f(x))\) (c) \(x h(f(x))\) (d) \(f(h(g(x)))\) (e) \(g(g(w))\) (f) \(h(h(t))\) (g) \(g(f(1 / x))\) (h) \(g(2 h(x-1))\) (i) Show that \(g(g(x)) \neq[g(x)]^{2}\) (j) Show that \([h(x)]^{2} \neq h\left(x^{2}\right)\).

Step-by-Step Solution

Answer

(a) \(1/x^2\) (b) \(1/(x^2)\) (c) \(3x^3 + x\) (d) \((3/x + 1)^2\) (e) \(w\) (f) \(9*t+4\) (g) \(x^2\) (h) \(1/(6x-1)\) (i) \(g(g(x)) \neq [g(x)]^{2}\) is shown (j) \([h(x)]^{2} \neq h(x^{2})\) is shown.

Step 1 - Find \(f(g(x))\)

Substitute \(g(x)=1 / x\) into \(f(x)\). This gives \(f(g(x))= [(1/x)]^2 = 1/x^2\)

Step 2 - Find \(g(f(x))\)

Substitute \(f(x)=x^{2}\) into \(g(x)\). This gives \(g(f(x)) = 1 / (x^2)\)

Step 3 - Find \(x h(f(x))\)

Substitute \(f(x)=x^{2}\) into \(h(x)\), then multiply the result with \(x\). This gives \(x h(f(x)) = x [3(x^2) +1] = 3x^3 + x\)

Step 4 - Find \(f(h(g(x)))\)

Substitute \(g(x) = 1/x\) into \(h(x)\), then substitute the result into \(f(x)\). This yields \(f(h(g(x))) = [3*(1/x)+1]^2 = (3/x + 1)^2\)

Step 5 - Find \(g(g(w))\)

Substitute \(g(x) = 1/x\) into itself. This yields \(g(g(w)) = 1 / (1/w) = w\)

Step 6 - Find \(h(h(t))\)

Substitute \(h(x) = 3x+1\) into itself. This yields \(h(h(t)) = 3*(3*t+1)+1 = 9*t+4\)

Step 7 - Find \(g(f(1 / x))\)

Substitute \(f(x) = x^2\) into \(g(x)\). This gives \(g(f(1/x)) = 1 / (1/x^2) = x^2\)

Step 8 - Find \(g(2 h(x-1))\)

Substitute \(h(x) = 3x+1\) into \(g(x)\), then multiply the result by 2. This yields \(g(2 h(x - 1)) = 1 / [3*(2*(x - 1)+1)+1] = 1/6x-1\)

Step 9 - Show that \(g(g(x)) \neq [g(x)]^{2}\)

Substitute \(g(x) = 1/x\) into itself to get \(g(g(x)) = w\), and square \(g(x)\) to get \([g(x)]^2 = 1/x^2\). As \(w \neq 1/x^2\), \(g(g(x)) \neq [g(x)]^{2}\) is shown

Step 10 - Show that \([h(x)]^{2} \neq h(x^{2})\)

Square \(h(x)\) to get \([h(x)]^2 = (3*x + 1)^2 = 9*x^2 + 6x + 1\) and substitute \(x^2\) into \(h(x)\) to get \(h(x^2) = 3*x^2 + 1\). As \(9*x^2 + 6x + 1 \neq 3*x^2 + 1\), \([h(x)]^{2} \neq h(x^{2})\) is shown

Key Concepts

Composite Functions

Understanding composite functions is like learning to play with nesting dolls—functions within functions, each one enclosing another. In mathematics, when you have two functions, say f(x) and g(x), you can create a new function by combining them. This process, called function composition, involves taking the output of one function and using it as the input for another.

For example, given f(x) = x^2 and g(x) = 1/x, the composite function f(g(x)) is found by substituting g(x) into f(x). Therefore, f(g(x)) = (1/x)^2. It’s important to work inside out, starting with the innermost function. This is the essence of creating composite functions—a crucial operation in advanced mathematics.

Functions and Their Rates of Change

In Calculus, one of the most vital concepts is the rate of change of a function. This is a measure of how a function's output changes as its input changes. To put it simply, it's like looking at how fast a car is going at any given point in time—it could be speeding up, slowing down, or cruising at a constant speed.

When dealing with composite functions, the rate of change becomes a little more intricate. You have to consider how each function within the composite function changes. If you look at the rate at which f(g(x)) changes with respect to x, you are in fact dealing with two rates: how f(x) changes with respect to g(x) and how g(x) changes with respect to x. The Chain Rule in Calculus is a technique which provides the ability to compute such rates for composite functions efficiently.

Calculus

Calculus is the branch of mathematics that deals with continuous change—think of it as the mathematics of motion and growth. It is divided into two main parts: differential calculus (concerned with rates of change and slopes of curves) and integral calculus (concerned with accumulation of quantities and the areas under and between curves).

The exercise provided is an excellent practical application of differential calculus, where you would use derivatives to find rates of changes in composite functions. Understanding how to work with functions and their compositions is fundamental, as many real-world phenomena can be described by functions and how quickly these functions change over time or with respect to other parameters.

Function Operations

In addition to composition, there are other operations which can be performed on functions including addition, subtraction, multiplication, and division. Much like arithmetic with numbers, these operations take two functions and combine them to make a new function. For example, if h(x) = 3x + 1, then the function h(h(t)) is a result of function operation, specifically a repeated composition, h(x) composed with itself.

Each operation on functions has its own rules, and understanding these rules is essential in solving complex mathematical problems. When functions interact through these operations, the possibilities are endless, broadening the landscape of mathematical solutions we can craft.