Problem 70 - College Algebra Essentials

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 simplified difference quotient of the function \(f(x)=-3 x^{2}+x-1\) is \(-6x - 3h + 1\).

Step 1: Substitute \(x + h\) into \(f(x)\)

The function is given as \(f(x) = -3x^{2} + x - 1\). Let's substitute \(x + h\) for \(x\) in the function: \(f(x+h) = -3(x + h)^{2} + (x + h) - 1\).

Step 2: Compute \(f(x + h)\)

We need to calculate this expression: \(f(x + h) = -3(x^{2} + 2xh + h^{2}) + x + h - 1 = -3x^{2} -6hx - 3h^{2} + x + h - 1\).

Step 3: Subtract \(f(x)\) from \(f(x + h)\)

Subtract \(f(x)\) from the computed \(f(x + h)\): \(\Delta f = f(x + h) - f(x) = -3x^{2} -6hx - 3h^{2} + x + h - 1 - (-3x^{2} + x - 1) = -6hx - 3h^{2} + h\).

Step 4: Divide by \(h\)

Now, divide this expression by \(h\): \(Quotient Difference = \frac{\Delta f}{h} = \frac{-6hx - 3h^{2} + h}{h} = -6x - 3h + 1\).

Key Concepts

Understanding Polynomial Functions

Polynomial functions are mathematical expressions that involve sums of powers of a variable, with coefficients as constants. In simpler terms, a polynomial is like a recipe where each ingredient is multiplied by a constant. The variable, usually represented by \(x\), can be raised to any non-negative integer power.
For example, in our given function, \(f(x) = -3x^2 + x - 1\), we have different 'terms':

• \(-3x^2\) is a quadratic term (as \(x\) is squared).
• \(x\) is a linear term (as \(x\) is to the first power).
• \(-1\) is a constant term.

Polynomial functions can take many shapes, often looking like curves or parabolas when graphed. In calculus, polynomial functions are key because they are smooth and continuous, making them ideal for analysis with calculus techniques. By understanding polynomials, we get a strong foundation in how different components interact and affect the function’s shape. This understanding is crucial when working with the difference quotient and eventually preparing for calculus.

The Art of Simplification

Simplification in mathematics is the process of making an expression as simple as possible. This means reducing it to fewer terms, or making those terms easier to interpret or work with. Let’s see how the concept of simplification is applied to our problem.
In step 2, for instance, after substituting \(x + h\) into \(f(x)\), we expand \(-3(x + h)^2 + (x + h) - 1\). This requires expanding \((x + h)^2\) to get \(x^2 + 2xh + h^2\), and then distributing the \(-3\) across the expanded terms:

• \(-3(x^2 + 2xh + h^2) = -3x^2 - 6xh - 3h^2\)
• Add \(x + h - 1\) to this:
• \(-3x^2 - 6xh - 3h^2 + x + h - 1\)

This expression is further simplified by combining like terms. Finally, when dividing by \(h\) in step 4, terms divisible by \(h\) are simplified, resulting in the difference quotient: \(-6x - 3h + 1\). This methodical reduction is essential because it helps reveal the underlying patterns and behaviors of the function, paving the way for calculus operations like differentiation.

Laying the Foundation for Calculus

The difference quotient is a crucial concept, especially in calculus preparation, because it lays the groundwork for understanding derivatives. The derivative is essentially the limit of the difference quotient as \(h\) approaches zero.
The quotient \(\frac{f(x+h)-f(x)}{h}\) is a measure of how the function value changes as we make a tiny shift in the variable \(x\). It's a way of computing the slope of the tangent line to the curve at any point. In practical terms, this gives an approximation of the point's instant rate of change.
Learning how to manipulate and simplify the difference quotient will help students understand what a derivative is and why it's useful. It deals with the concept of instantaneous change, which is fundamental in physics, engineering, and economics, among many fields. Grasping these ideas not only prepares you for calculus but also empowers you to solve real-world problems where change is a crucial factor.