Problem 62 - Intermediate Algebra

Question

Find \(\frac{f(a+h)-f(a)}{h}\) for each of the given functions. (Objective 4) $$f(x)=x^{2}-3 x+7$$

Step-by-Step Solution

Answer

\(\frac{f(a+h)-f(a)}{h} = 2a + h - 3.\)

Step 1: Substitute x with a+h

The function given is \( f(x) = x^2 - 3x + 7 \). We need to substitute \( x \) with \( a+h \) to find \( f(a+h) \). This yields:\[f(a+h) = (a+h)^2 - 3(a+h) + 7.\]

Step 2: Expand f(a+h)

Next, we expand \( f(a+h) \). This involves squaring \( a+h \) and distributing the \( -3 \) across \( (a+h) \):\[(a+h)^2 = a^2 + 2ah + h^2\]\[f(a+h) = a^2 + 2ah + h^2 - 3a - 3h + 7.\]

Step 3: Substitute x with a

Substitute \( x \) with \( a \) in the original function to find \( f(a) \):\[f(a) = a^2 - 3a + 7.\]

Step 4: Compute f(a+h) - f(a)

Calculate \( f(a+h) - f(a) \) by subtracting the expression for \( f(a) \) from \( f(a+h) \):\[f(a+h) - f(a) = (a^2 + 2ah + h^2 - 3a - 3h + 7) - (a^2 - 3a + 7).\]Simplify:\[f(a+h) - f(a) = 2ah + h^2 - 3h.\]

Step 5: Divide by h

Finally, divide the expression \( f(a+h) - f(a) \) by \( h \):\[\frac{f(a+h)-f(a)}{h} = \frac{2ah + h^2 - 3h}{h}.\]Simplify the expression by factoring out \( h \):\[\frac{f(a+h)-f(a)}{h} = 2a + h - 3.\]

Key Concepts

Polynomial Functions

Polynomial functions are one of the most significant classes of functions in algebra. They consist of monomials which are terms composed of variables raised to whole number powers. The function's standard form is written as

• \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \)

Where

• \( a_n, a_{n-1}, \ldots, a_1, a_0 \) are constants called coefficients.
• \( n \) is a non-negative integer, known as the degree of the polynomial.

In our exercise, the function \( f(x) = x^2 - 3x + 7 \) is a polynomial of degree 2 because the highest power of \( x \) is 2. Polynomial functions can describe a wide range of real-world phenomena and are therefore found frequently in various fields of science and engineering. Their importance in calculus, especially, stems from the fact that they are smooth and continuous with well-behaved derivatives.

Algebraic Expressions

Algebraic expressions involve the combination of variables, constants, and operations like addition, multiplication, and exponentiation. Understanding these expressions is crucial, as they form the backbone of how we solve problems in algebra and calculus.When expanding expressions such as \((a+h)^2\), you perform a series of operations based on algebraic rules. By applying the distributive property, you expand \((a+h)^2 = a^2 + 2ah + h^2\). Each term in the expanded form comes from applying multiplication over the binomial.In our solution, after substituting \( a+h \) into the polynomial, they expand the expression to simplify and prepare for function evaluation. Arriving at the expanded form, \( f(a+h) = a^2 + 2ah + h^2 - 3a - 3h + 7 \), is crucial for correctly setting up the difference quotient calculation that follows.

Function Evaluation

Function evaluation involves substituting specific values into the function to calculate numerical outputs. It's a fundamental operation in mathematics, enabling us to examine how a function behaves under particular conditions.In the original problem, you evaluate the function \( f(x) \) by substituting \( x \) with different values like \( a \) to find \( f(a) \), and with \( a+h \) to find \( f(a+h) \). The key step in this process involves calculating \( f(a+h) - f(a) \), which, after simplification, becomes \( 2ah + h^2 - 3h \). This expression is vital in the context of the difference quotient, traditionally used to understand changes in the function's behavior—particularly in preparation for calculus concepts such as derivatives.By dividing this expression by \( h \), we simplify it further to \( 2a + h - 3 \), representing how the function changes as \( h \) approaches zero. This simplification is essential for deeper calculus and algebra applications, providing insights into the function's rate of change at \( x = a \).