Problem 56 - Contemporary Precalculus

Question

Find \(g\) o \(f\) and find the difference quotient of the function \(g \circ f\) $$f(x)=x^{3} ; g(x)=x+2$$

Step-by-Step Solution

Answer

Answer: The difference quotient of the composition of functions g(x) and f(x) is $$3x^2 + 3xh + h^2$$.

Step 1: Find the composition of g and f

First, we need to find the composition of the functions g(x) and f(x). This can be done by plugging f(x) into g(x): \((g \circ f)(x) = g(f(x))\). $$ (g \circ f)(x) = g(x^3)= (x^3) +2 $$ So, \((g \circ f)(x) = x^3 + 2\).

Step 2: Calculate the difference quotient

Now that we have the composition, we can find the difference quotient. Using the formula mentioned in the analysis: $$\frac{g\circ f(x + h) - g\circ f(x)}{h}$$ We can substitute the function \((g \circ f)(x) = x^3 + 2\) into the formula: $$\frac{((x + h)^3 + 2) - (x^3 + 2)}{h}$$

Step 3: Simplify the difference quotient

We can now simplify the expression by expanding and canceling terms: $$\frac{((x + h)^3 + 2) - (x^3 + 2)}{h} = \frac{(x^3 + 3x^2h + 3xh^2 + h^3 + 2) - (x^3 + 2)}{h}$$ Cancel the \(x^3\) and \(2\) terms: $$\frac{3x^2h + 3xh^2 + h^3}{h}$$ Since all of the terms in the numerator have h as a factor, we can factor out h: $$\frac{h(3x^2 + 3xh + h^2)}{h}$$ Now, we can cancel out the h term from the numerator and the denominator: $$3x^2 + 3xh + h^2$$ So, the difference quotient of the function \((g \circ f)(x)\) is: $$3x^2 + 3xh + h^2$$

Key Concepts

Understanding the Difference Quotient

The difference quotient is a crucial concept within calculus because it represents the average rate of change of a function over a specific interval. More specifically, it is the slope of the secant line between two points on a curve. This is the basis for understanding derivatives.
To calculate it, we use the formula:

• \( \frac{f(x + h) - f(x)}{h} \)

In the context of our exercise, where we have a composed function \((g \circ f)(x) = x^3 + 2\), the difference quotient becomes:
\[ \frac{((x + h)^3 + 2) - (x^3 + 2)}{h} \]
This formula helps us approximate the instantaneous rate of change, preparing us for more advanced calculus concepts such as derivatives.

Exploring Polynomial Functions

Polynomial functions are expressions involving variables and coefficients, constructed from operations of addition, subtraction, multiplication, and non-negative integer exponents. An example from our exercise is \(f(x) = x^3\), a simple polynomial of degree 3.
When you compose functions, such as when we combine \(f(x)\) and \(g(x)\) in our example to form \((g \circ f)(x) = x^3 + 2\), we still maintain a polynomial structure. It's notable how adding a simple constant term, like 2, influences the output without altering the degree of the polynomial.
Understanding polynomials help students in calculus since many real-world problems can be approximated by these functions easing further analysis.

Connecting to Calculus Concepts

The exercise of finding the difference quotient and function composition ties directly into fundamental calculus concepts. Comprehension of these concepts is vital for understanding limits and the derivative process.
When computing \(g \circ f\), we were performing what's called function composition. This is where one function is applied to the results of another. In calculus, mastering this technique allows us to tackle more intricate problems involving composite functions.
Moreover, the simplification of the difference quotient from lengthy expressions to \(3x^2 + 3xh + h^2\) reflects the calculus practice of finding derivatives. As calculus progresses, these foundational skills foster the precise calculation of rates of change and slopes of curves, becoming essential tools for advanced studies.