Problem 31 - Contemporary Precalculus

Question

Let \(f(x)=x^{2}+5,\) and let \(g(x)=f(x-1)\) (a) Write the rule of \(g(x)\) and simplify. (b) Find the difference quotients of \(f(x)\) and \(g(x)\) (c) Let \(d(x)\) denote the difference quotient of \(f(x) .\) Show that the difference quotient of \(g(x)\) is \(d(x-1)\)

Step-by-Step Solution

Answer

Question: Show that the difference quotient of the function \(g(x) = x^2 - 2x + 6\) based on the function \(f(x) = x^2 + 5\) is d(x-1) where d(x) denotes the difference quotient of \(f(x)\). Answer: The difference quotient of \(g(x)\) is \(2x + h - 2\), and the difference quotient of \(f(x)\) is \(2x + h\). By replacing x with \((x-1)\) in \(d(x)\), we get \(d(x-1) = 2(x-1) + h = 2x - 2 + h\). Since \(g'(x) = d(x-1)\), the difference quotient of \(g(x)\) is indeed \(d(x-1)\).

Step 1: Find the rule of G(x) and simplify

To find the rule of \(g(x)\), we will substitute \((x-1)\) into \(f(x)\). So, \(g(x) = f(x-1) = (x-1)^2 + 5\). Now let's simplify this: \(g(x) = (x^2 - 2x + 1) + 5\) \(g(x) = x^2 - 2x + 6\) So, the rule of \(g(x)\) is \(x^2 - 2x + 6\).

Step 2: Find the difference quotients of F(x)

The difference quotient of a function is defined as: \(\frac{f(x+h) - f(x)}{h}\), where \(h \neq 0\). Let's find the difference quotient of \(f(x)\): \(\frac{f(x+h) - f(x)}{h} = \frac{((x+h)^2+5) - (x^2 + 5)}{h}\) \( = \frac{x^2 + 2xh + h^2 + 5 - x^2 - 5}{h}\) \( = \frac{2xh + h^2}{h}\) Now we can factor out h: \( = h(2x+h)/h\) \(h\) cancels from the numerator and denominator: \( = 2x + h\) So, the difference quotient of \(f(x)\) is \(2x + h\).

Step 3: Find the difference quotients of G(x)

Now let's find the difference quotient of \(g(x)\): \(\frac{g(x+h) - g(x)}{h} = \frac{[(x+h)^2 - 2(x+h) + 6] - (x^2 - 2x + 6)}{h}\) \( = \frac{x^2 + 2xh + h^2 - 2x - 2h + 6 - x^2 + 2x - 6}{h}\) \( = \frac{2xh + h^2 - 2h}{h}\) Now we can factor out h: \( = h(2x + h - 2)/h\) \(h\) cancels from the numerator and denominator: \( = 2x + h- 2\) So, the difference quotient of \(g(x)\) is \(2x + h -2\).

Step 4: Show that the difference quotient of G(x) is d(x-1)

We have found that the difference quotient of \(f(x)\) is \(2x + h\), and the difference quotient of \(g(x)\) is \(2x + h - 2\). Recall that \(d(x)\) denotes the difference quotient of \(f(x)\). By replacing x with \((x-1)\) in \(d(x)\), we get \(d(x-1) = 2(x-1) + h = 2x - 2 + h\). Comparing this with the difference quotient of \(g(x)\), we have: \(g'(x) = 2x + h - 2 = d(x-1)\) Therefore, the difference quotient of \(g(x)\) is \(d(x-1)\).

Key Concepts

Function Transformation

Function transformations deal with changing a function's formula to achieve shifts or distortions in its graph. These transformations include translations, dilations, reflections, and rotations.
For this problem, we are focused on a translation. Specifically, we take the function \(f(x) = x^2 + 5\) and substitute \((x-1)\) to form \(g(x) = f(x-1)\). This action shifts the graph of \(f(x)\) horizontally by 1 unit to the right.
By simplifying \(f(x-1)\), we calculate:

• Substitute \((x-1)\) into \(f(x)\): \((x-1)^2 + 5\).
• Expand \((x-1)^2\) to obtain \(x^2 - 2x + 1\).
• Therefore, \(g(x) = x^2 - 2x + 1 + 5 = x^2 - 2x + 6\).

The resulting function is \(g(x) = x^2 - 2x + 6\), signifying that it has been transformed through a horizontal shift.

Polynomial Functions

Polynomial functions, like \(f(x) = x^2 + 5\), are expressions involving a combination of powers of \(x\) with coefficients. Here we focus on quadratic polynomial functions, which are degree 2. These are common in mathematics due to their parabolic graph shapes.
Understanding polynomials involves recognizing:

• Degree: The highest power of \(x\).
• Coefficients: Numbers preceding terms like \(a\) in \(ax^2\).
• Constants: Terms without an \(x\), like the \(+5\) in \(x^2 + 5\).

When transforming functions, such as \(g(x) = f(x-1)\), observe changes in coefficients and constants to understand transformations on the graph. Here, the distribution of \(-2x\) indicates a horizontal shift has occurred.

Mathematical Proof

Mathematical proof involves demonstrating the truth of a statement using logical reasoning. In this exercise, we show that two difference quotients relate by proving they match when a certain transformation is applied.
Steps in this proof include:

• Calculate the difference quotient for both \(f(x)\) and \(g(x)\).
• For \(f(x)\), derive \(\frac{f(x+h) - f(x)}{h} = 2x + h\).
• For \(g(x)\), find \(\frac{g(x+h)- g(x)}{h} = 2x + h - 2\).
• Substitute \((x-1)\) in \(d(x)\) for \(x\) to get \(d(x-1) = 2x - 2 + h\).

Finally, compare terms to conclude \(g'(x) = 2x + h - 2 = d(x-1)\). This verifies the transformation linked \(g(x)\) to \(d(x-1)\).

Algebraic Simplification

Algebraic simplification helps in making expressions more manageable by reducing them to simpler forms or factorizing. Here, our main task was simplifying the function \(g(x) = (x-1)^2 + 5\).
Steps involved:

• Simplifying \((x-1)^2\) by expanding to \(x^2 - 2x + 1\).
• Adding the constant \(+5\) to get \(x^2 - 2x + 6\).
• Check for common factors or substitution errors.

Such operations are vital when dealing with polynomial transformations and proofs as they lay out a clearer path to understanding concepts. Simplification reduces complexity and allows us to observe the core transformations and their effects more clearly.