Problem 61 - Intermediate Algebra

Question

Find \(\frac{f(a+h)-f(a)}{h}\) for each of the given functions. (Objective 4) $$f(x)=-x^{2}-1$$

Step-by-Step Solution

Answer

\( \frac{f(a+h)-f(a)}{h} = -2a - h \).

Step 1: Substitute in the function definition

First, identify the given function, which is \( f(x) = -x^2 - 1 \). We need to evaluate \( f(a+h) \) and \( f(a) \). Substitute \( x = a+h \) into the function to get \( f(a+h) = -(a+h)^2 - 1 \).

Step 2: Expand \( f(a+h) \)

Now we need to expand the expression for \( f(a+h) \). Calculate \((a+h)^2\) which gives \(a^2 + 2ah + h^2\). Substitute it back into the function to get \(-((a+h)^2) = -(a^2 + 2ah + h^2) = -a^2 - 2ah - h^2\). So, \(f(a+h) = -a^2 - 2ah - h^2 - 1\).

Step 3: Substitute and simplify \( f(a) \)

For \( f(a) \), directly substitute \( a \) into the function: \( f(a) = -a^2 - 1 \).

Step 4: Calculate \( \frac{f(a+h)-f(a)}{h} \)

Now, use the expressions found in the previous steps: \( f(a+h) = -a^2 - 2ah - h^2 - 1 \) and \( f(a) = -a^2 - 1 \).Subtract \( f(a) \) from \( f(a+h) \):\[(-a^2 - 2ah - h^2 - 1) - (-a^2 - 1) = -2ah - h^2\]Divide the result by \( h \):\[\frac{-2ah - h^2}{h} = -2a - h\].

Step 5: Simplify the expression

We have found that \( \frac{f(a+h)-f(a)}{h} = -2a - h \). This is the simplified final answer for the given function.

Key Concepts

Quadratic Functions

Quadratic Functions are a special type of polynomial functions where the highest degree of any term is two. They generally take the form:

• \( f(x) = ax^2 + bx + c \)

Here, \( a, b, \) and \( c \) are constants, and \( a eq 0 \). Quadratic functions often have the iconic U-shaped curve called a parabola. This curve can open upwards or downwards, depending on the sign of \( a \). When \( a > 0 \), the parabola opens upwards; when \( a < 0 \), it opens downwards.
In the given exercise, the function \( f(x) = -x^2 - 1 \) is quadratic.
The negative sign in front of \( x^2 \) indicates that the parabola faces downwards. Understanding these properties can help you visualize and predict how changes to a quadratic function's parameters will affect its graph.

Function Substitution

Function Substitution involves replacing a variable in a function with another expression. It's a fundamental concept in calculus and algebra. In this context, it allows you to evaluate a function at a specific point or expression. For example:

• If you have \( f(x) = -x^2 - 1 \) and want to find \( f(a+h) \), you substitute \( x \) with \( a+h \).

This gives you:

• \( f(a+h) = -(a+h)^2 - 1 \)

By applying this substitution method, we can derive explicit expressions that reveal detailed insights about a function's behavior when input changes. Simplifying \( (a+h)^2 \) to \( a^2 + 2ah + h^2 \) helps in expanding the function correctly for further calculations.

Algebraic Simplification

Algebraic Simplification is the process of reducing an expression to its most basic form while preserving its value. This typically involves combining like terms, removing parentheses, and reducing fractions. In our exercise, we started with:

• \( f(a+h) = -a^2 - 2ah - h^2 - 1 \)
• \( f(a) = -a^2 - 1 \)

To simplify \( \frac{f(a+h) - f(a)}{h} \), the original expressions are substituted and simplified:

• Subtract \( f(a) \) from \( f(a+h) \) which becomes: \(-2ah - h^2\).
• Divide by \( h \): \( \frac{-2ah - h^2}{h} = -2a - h \).

This process of algebraic simplification is essential to effectively understanding and solving algebraic problems. It can also be a stepping stone to more advanced topics like calculus.