Problem 66 - College Algebra

Question

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

Step-by-Step Solution

Answer

The simplified difference quotient is \(-2x - 3\).

Step 1: Apply the Difference Quotient Formula

Substitute \(f(x)\) and \(f(x+h)\) in the formula. \(f(x) = -x^{2}-3x+1\) and \(f(x+h) = -(x+h)^{2} - 3(x+h) + 1\). The difference quotient is \(\frac{-((x+h)^{2} + 3(x+h) - 1) - (-x^{2}-3x+1)}{h}\)

Step 2: Simplify the Numerator

Expand and simplify the numerator: \( -((x+h)^{2} + 3(x+h) - 1) - (-x^2 -3x +1) = -x^{2} - 2hx - h^{2} - 3x - 3h + 1 - (-x^{2}-3x+1) = -2hx - h^{2} - 3h\)

Step 3: Simplify the Difference Quotient

Finally, simplify the difference quotient by cancelling out the h in the numerator with the h in the denominator: \(\frac{-2hx - h^2 - 3h}{h} = -2x - h - 3\)

Step 4: Final Simplification

The difference quotient as h approaches 0 gives the derivative. If h approaches 0, then \(-2x - h - 3\) approximately simplifies to \(-2x - 3\). This is the simplified form of the difference quotient.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial characterized by the squared term, or degree of two. In general, quadratic functions can be expressed in the form \( f(x) = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \) ensures that the function has a parabolic shape.
For the given exercise, the quadratic function is \( f(x) = -x^2 - 3x + 1 \). Each term contributes differently:

• The \( -x^2 \) makes the parabola open downwards.
• The \( -3x \) determines the slope of the tangent at different points along the curve.
• The constant \( +1 \) shifts the entire parabola up one unit on the y-axis.

Understanding these parts helps in predicting the overall shape and the direction of the graph for any quadratic function. This foundation plays a significant role when we delve into calculus concepts like the difference quotient.

Simplifying Expressions

Simplifying expressions involves breaking down complex mathematical expressions into simpler, more digestible forms. This process is key when working with the difference quotient in calculus.
To simplify expressions effectively, follow these steps:

• Expand all products. For example, \((x+h)^2\) expands to \(x^2 + 2xh + h^2\).
• Combine like terms, such as merging all \(x^2\), \(x\), and constant terms.
• Factor out common terms to simplify complex parts of the expression.

In our exercise, after applying the difference quotient formula, the result was simplified by expanding and combining like terms in the numerator. This gave the expression \(-2hx - h^2 - 3h\), which could then eliminate \( h \) when divided by it. Simplifying rules like these reduce errors and make calculations more manageable.

Calculus Concepts

Calculus introduces many vital concepts, one of which is the difference quotient. It's a key tool in understanding how functions change, specifically leading to the concept of the derivative. The difference quotient is expressed as \( \frac{f(x+h)-f(x)}{h} \) and essentially finds the average rate of change of the function over an interval.
In the context of the quadratic function \( f(x) = -x^2 - 3x + 1 \), the difference quotient was used to determine how the function changes as we move along its curve.

• Step 1: Write \( f(x+h) \) by replacing \( x \) with \( x+h \) in the function.
• Step 2: Plug these into the quotient and simplify systematically.
• Step 3: Cancel out terms to approach the derivative as \( h \) nears zero.

Once \( h \) approaches zero, the expression \(-2x - h - 3\) simplifies to \(-2x - 3\), which represents the derivative of \( f(x) \). This gives the instantaneous rate of change and is foundational for further calculus studies.