Problem 70 - College Algebra

Question

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$ for the given function. $$f(x)=-3 x^{2}+x-1$$

Step-by-Step Solution

Answer

The difference quotient for the function \(f(x) = -3x^{2}+x-1\) simplifies to \(-6x -3h + 1\).

Step 1: Substitute \(x+h\) into the function

Replace x in the function \(f(x)=-3x^{2}+x-1\) with \(x+h\). This gives us the following expression: \(f(x+h) = -3(x+h)^{2}+(x+h)-1 = -3(x^{2}+2xh + h^{2}) + x + h -1\).

Step 2: Distribute the -3 to terms inside of the parantheses

After distributing, we get \(f(x+h) = -3x^{2} - 6xh -3h^{2} + x + h -1\).

Step 3: Subtract the function

Now subtract the given function \(f(x)=-3x^{2}+x-1\) from \(f(x+h) = -3x^{2} - 6xh - 3h^{2} + x + h -1\). So the computation to perform is \(( -3x^{2} - 6xh - 3h^{2} + x + h -1) - ( -3x^{2}+x-1)\). After subtracting, this gives us \(-6xh -3h^{2} + h\).

Step 4: Divide by h

Finally, divide the whole expression by h. Thus, the quantity \(-6xh -3h^{2} + h\) divided by h simplifies to \(-6x -3h + 1\). This is the difference quotient for the given function.

Key Concepts

Function Substitution

Function substitution is a crucial step in calculus that makes working with functions easier. We replace the variable(s) in a function with a given expression to find the resultant function. In this exercise, we substitute \(x+h\) into the original function \(f(x) = -3x^{2} + x - 1\). This process transforms the function into \(f(x+h)\), giving us an expression in terms of \(x\) and \(h\) instead of just \(x\). Substitution is straightforward yet powerful, as it sets the stage for further operations like simplification and evaluation.

To substitute:

• Identify the variable in the function, typically represented by \(x\).
• Replace each occurrence of the variable with the new expression \(x+h\).
• Ensure all instances of the variable are replaced consistently for accurate results.

This step might appear simple, but it’s a crucial foundation for dealing with more complex calculus concepts.

Polynomial Functions

Polynomial functions, like \(f(x) = -3x^{2} + x - 1\), are functions composed of terms derived from constants and variables raised to non-negative integer powers. These functions are quintessential in mathematics because of their stability and the role they play in various applications. Understanding their structure helps tremendously in tasks like substitution, differentiation, or integration.

A polynomial's degree is the highest power of the variable in the function, dictating the shape and complexity of its graph. In our example, the degree is 2, making it a quadratic function. Here are some features of polynomial functions:

• Terms are composed of coefficients (like \(-3, 1, -1\)) and the variables \(x\).
• They can have multiple terms but a single output value for any given input.
• Polynomial functions are continuous and smooth over all real numbers, enabling predictions and calculations over different domains.

Recognizing these aspects of polynomial functions assists in simplifying and analyzing expressions efficiently.

Algebraic Simplification

Once substitutions are made, simplifying the resulting expression is key to making calculations manageable and understandable. Algebraic simplification involves combining like terms and minimizing the complexity of expressions using fundamental algebra rules.

In simplifying the difference quotient of our function, we first expanded \(-3(x+h)^2\) to turn the substitution into an explicit form. The primary goal was to reflect all terms linearly or as a simple polynomial to allow smooth continuation of mathematical operations. Here’s how simplification is tackled:

• Distribute coefficients through parentheses to remove nested expressions, as shown with the \(-3\) multiplier.
• Combine like terms to reduce the polynomial degree when possible, such as merging \(-6xh\) and others.
• Cancel redundant elements, especially when dividing by variables like \(h\), to simplify calculations to terms like \(-6x - 3h + 1\).

Each step ensures we avoid unnecessarily complex expressions, turning algebraic endeavors from daunting to doable.