Problem 47 - Calculus and its applications

Question

Find the simplified difference quotient for each function listed. $$ f(x)=a x^{2}+b x+c $$

Step-by-Step Solution

Answer

The simplified difference quotient is \(2ax + ah + b\).

Step 1: Express the Difference Quotient

The difference quotient for a function \(f(x)\) is given by\[\frac{f(x + h) - f(x)}{h}\]We need to evaluate \(f(x + h)\) based on the function \(f(x) = ax^2 + bx + c\).

Step 2: Substitute and Expand \(f(x + h)\)

First, substitute \(x + h\) into the function:\[f(x + h) = a(x + h)^2 + b(x + h) + c\]Now, expand \((x + h)^2\) to get:\[(x + h)^2 = x^2 + 2xh + h^2\]Expand the entire function:\[f(x + h) = a(x^2 + 2xh + h^2) + b(x + h) + c\]This becomes\[ax^2 + 2axh + ah^2 + bx + bh + c\]

Step 3: Compute \(f(x + h) - f(x)\)

Substitute the expression for \(f(x + h)\) and \(f(x)\):\[(f(x + h) - f(x)) = (ax^2 + 2axh + ah^2 + bx + bh + c) - (ax^2 + bx + c)\]Simplify by cancelling out like terms:\[= 2axh + ah^2 + bh\]

Step 4: Divide by \(h\)

Now, substitute this result back into the difference quotient formula:\[\frac{2axh + ah^2 + bh}{h}\]Simplify by dividing each term by \(h\):\[2ax + ah + b\]

Step 5: Simplified Difference Quotient

The simplified difference quotient for the function \(f(x) = ax^2 + bx + c\) is:\[2ax + ah + b\]

Key Concepts

Subtracting Polynomials

When working with polynomials, it's important to understand how to subtract one from another. Polynomials are algebraic expressions that include terms like \(ax^2\), \(bx\), and \(c\). Subtracting polynomials involves taking each corresponding term of two polynomial expressions and performing the subtraction operation.

This step is particularly important when dealing with difference quotients, as you'll often subtract one polynomial expression from another.

Here's a quick guide to subtract polynomials:

• Arrange the polynomials in a vertical format if helpful, aligning like terms vertically.
• Subtract each pair of like terms (i.e., terms with the same degree).
• Remember to distribute the negative sign to all terms of the polynomial being subtracted.
• Combine the results for a final simplified subtraction result.

Using these steps helps ensure that when you find the difference between \(f(x+h)\) and \(f(x)\), you can properly cancel and subtract terms like \(ax^2\), \(bx\), and \(c\), as seen in our original solution.

Expanding Expressions

Expanding expressions is a crucial skill when working with polynomials and algebraic expressions. This process generally involves taking an expression written in a compact or factored form and rewriting it in an expanded form, especially when taking powers of binomials like \((x+h)^2\).

For the function \(f(x) = ax^2 + bx + c\), expanding \((x + h)^2\) is a prime example of this process. Let's break it down:

• Start by expanding the binomial \((x + h)^2 = x^2 + 2xh + h^2\).
• Multiply each term in this expanded binomial with the respective coefficients from the function, in this case, multiply by \(a\).
• Continue to distribute and expand any additional terms in the expression, such as \(b(x + h)\), to yield \(bx + bh\).

By expanding the expression fully, you can then substitute it back into the difference quotient formula and simplify accordingly.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is an essential part of calculus and algebra, especially when finding the difference quotient. Simplification involves reducing an expression to its simplest possible form, making it easier to analyze or solve.

Once you've expanded and set up your expressions, you will need to simplify them to find usable results. Here's how you can do it:

• First, identify and cancel out like terms that appear in both expressions - such as \(ax^2\), \(bx\), and \(c\) from our example.
• Add or subtract coefficients for like terms to combine them into a single, simplified expression.
• Factor out common factors where possible to make expressions cleaner and more compact.

Simplifying is particularly important in problems involving the difference quotient because it allows you to see the primary factors affecting the change, such as \(2ax + ah + b\) for the given polynomial function.