Problem 8 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 5-10: If \(a\) and \(h\) are real numbers, find (a) \(f(a)\) (b) \(f(-a)\) (c) \(-f(a)\) (d) \(f(a+h)\) (e) \(f(a)+f(h)\) (f) \(\frac{f(a+h)-f(a)}{h}\), if \(h \neq 0\) $$ f(x)=3-x^{2} $$

Step-by-Step Solution

Answer

(a) \( f(a) = 3 - a^2 \); (b) \( f(-a) = 3 - a^2 \); (c) \( -f(a) = a^2-3 \); (d) \( f(a+h) = 3 - a^2 - 2ah - h^2 \); (e) \( f(a)+f(h) = 6-a^2-h^2 \); (f) \( \frac{-2ah-h^2}{h} = -2a-h \).

Step 1: Find \( f(a) \)

To find \( f(a) \), substitute \( a \) for \( x \) in the function definition. Thus, \( f(a) = 3 - a^2 \).

Step 2: Find \( f(-a) \)

Replace \( x \) with \( -a \) in the function. Hence, \( f(-a) = 3 - (-a)^2 = 3 - a^2 \) because \((-a)^2 = a^2\).

Step 3: Find \( -f(a) \)

First, determine \( f(a) \) which is \( 3 - a^2 \). Then, the negative of this is \( -f(a) = -(3 - a^2) = a^2 - 3 \).

Step 4: Find \( f(a+h) \)

Substitute \( a+h \) for \( x \) in \( f(x) \). This yields \( f(a+h) = 3 - (a+h)^2 = 3 - (a^2 + 2ah + h^2) \), which simplifies to \( 3 - a^2 - 2ah - h^2 \).

Step 5: Find \( f(a) + f(h) \)

Calculate \( f(a) = 3 - a^2 \) and \( f(h) = 3 - h^2 \). Add these results: \( f(a) + f(h) = (3 - a^2) + (3 - h^2) = 6 - a^2 - h^2 \).

Step 6: Find \( \frac{f(a+h) - f(a)}{h} \) when \( h \neq 0 \)

Subtract \( f(a) \) from \( f(a+h) \): \( (3 - a^2 - 2ah - h^2) - (3 - a^2) = -2ah - h^2 \). Now, divide by \( h \): \( \frac{-2ah - h^2}{h} = -2a - h \) (since \( h eq 0 \)).

Key Concepts

Polynomial Functions

Polynomial functions are incredibly versatile. They involve expressions that are composed of variables raised to whole number powers and multiplied by coefficients. A simple polynomial function is given by \( f(x) = 3 - x^2 \). Here, this function is quadratic because the highest power of \( x \) is 2, making it a parabola when graphed on the coordinate plane. The expression "3 minus \( x^2 \)" immediately suggests a polynomial of degree 2, indicating a parabola that opens downwards, as seen in the term \(-x^2\).
Studying polynomial functions involves understanding their structure and behavior. You can determine:

• The degree, which provides insights into the shape and number of turning points.
• The coefficients, which influence the stretch or compression of the graph.
• The constant term, which indicates where the graph intersects the y-axis.

Being comfortable with evaluating polynomial functions for specific values, like substituting \( a \) or \( a+h \), helps in understanding how these functions change in different contexts.

Algebra

Algebra is the language of mathematics. It allows us to express relationships through symbols, like letters representing numbers. For example, in the given function \( f(x) = 3 - x^2 \), we use \( x \) to denote any number. An important algebraic skill is substituting different values into the function.
In step 1, we found \( f(a) \) by replacing \( x \) with \( a \). This yields \( 3 - a^2 \). Algebra handles such substitutions with ease. In algebra, we also manipulate expressions. For instance, finding \( f(a+h) \) involves expanding the expression \((a+h)^2\) to reveal more terms: \( a^2 + 2ah + h^2 \).
Key algebraic processes include:

• Substitution: Inserting specific values for variables.
• Expansion: Distributing and combining like terms.
• Simplification: Reducing expressions to their simplest form.

Mastering these skills ensures you can tackle a range of problems involving polynomial functions and beyond.

Difference Quotient

The difference quotient is a fundamental concept in calculus, representing the average rate of change of a function over an interval. This concept often prepares students for understanding derivatives. In the exercise, we deal with the difference quotient in part (f): \( \frac{f(a+h) - f(a)}{h} \).
Calculating this involves the following:

• First, substitute and find \( f(a+h) \), which is \( 3 - a^2 - 2ah - h^2 \).
• Next, subtract \( f(a) \) (\( 3 - a^2 \)) from \( f(a+h) \), to get \( -2ah - h^2 \).
• Then, divide this result by \( h \), provided \( h eq 0 \), yielding the simplified form \( -2a - h \).

The difference quotient gives insight into how the function \( f(x) \) changes as \( x \) changes from \( a \) to \( a+h \). It emphasizes concepts of slope and change, foundational in calculus where we move towards finding exact rates of change at any point.