Problem 41 - Algebra and Trigonometry

Question

Find \(f(a), f(a+h),\) and the difference quotient \(\frac{f(a+h)-f(a)}{h},\) where \(h \neq 0\) $$ f(x)=3-5 x+4 x^{2} $$

Step-by-Step Solution

Answer

\( f(a) = 3 - 5a + 4a^2 \), \( f(a+h) = 3 - 5a - 5h + 4a^2 + 8ah + 4h^2 \), difference quotient: \(-5 + 8a + 4h\).

Step 1: Evaluate f(a)

To find \( f(a) \), substitute \( a \) for \( x \) in the function \( f(x) = 3 - 5x + 4x^2 \). Thus, \( f(a) = 3 - 5a + 4a^2 \).

Step 2: Evaluate f(a+h)

Substitute \( a+h \) for \( x \) in the function \( f(x) = 3 - 5x + 4x^2 \).\\( f(a+h) = 3 - 5(a+h) + 4(a+h)^2 \).\Now, expand \( (a+h)^2 \) to obtain \( a^2 + 2ah + h^2 \).\Thus, \( f(a+h) = 3 - 5a - 5h + 4(a^2 + 2ah + h^2) \).\This simplifies to \( f(a+h) = 3 - 5a - 5h + 4a^2 + 8ah + 4h^2 \).

Step 3: Calculate the Difference Quotient

The difference quotient is given by \( \frac{f(a+h) - f(a)}{h} \). Substitute the expressions from Steps 1 and 2: \[\frac{(3 - 5a - 5h + 4a^2 + 8ah + 4h^2) - (3 - 5a + 4a^2)}{h}\]Simplify the numerator: the \( 3 \), \(-5a\), and \( 4a^2 \) terms cancel out, leaving:\[-5h + 8ah + 4h^2\].Factor \( h \) out of the numerator: \( h(-5 + 8a + 4h) \).Thus, the difference quotient simplifies to:\[-5 + 8a + 4h\].

Key Concepts

Polynomial Functions

Polynomial functions are expressions consisting of variables and coefficients, structured in a way where operations include addition, subtraction, multiplication, and non-negative integer exponents. They are fundamental in algebra and help describe various types of natural phenomena. In our given exercise, the function \(f(x) = 3 - 5x + 4x^2\) is a polynomial. Its degree is determined by the highest power of \(x\), which in this case is 2 (from \(4x^2\)), making it a quadratic polynomial.

• Linear Polynomial: Consists of terms up to the first degree. Example: \(ax + b\).
• Quadratic Polynomial: Involves terms up to the second degree. Example: \(ax^2 + bx + c\).
• Cubic Polynomial: Includes terms up to the third degree. Example: \(ax^3 + bx^2 + cx + d\).

Polynomial functions often contain interaction between terms, like the "cross-terms" seen when expanding expressions like \((a+h)^2\). Understanding these interactions is key in solving problems involving polynomials, such as evaluating the difference quotient in this exercise.

Function Evaluation

Function evaluation involves finding the output of a function given specific inputs. This requires substituting values into a function's equation. In the context of our exercise, you evaluate \(f(a)\) by substituting \(a\) for \(x\) in the polynomial function \(f(x) = 3 - 5x + 4x^2\), yielding \(f(a) = 3 - 5a + 4a^2\). Similarly, for \(f(a+h)\), we replace \(x\) with \(a+h\), leading to more complex expressions due to the expanded terms.

• Evaluate with direct substitution: Often involves plugging in a given number or expression into each occurrence of the variable.
• Recognize the effect of change: Altering the variable by adding \(h\) results in additional terms that reveal how the function changes incrementally.

By systematically substituting and simplifying, you can determine a polynomial's behavior and make predictions about its values, which is essential when working with function evaluations like \(f(a)\) and \(f(a+h)\).

Algebraic Expressions

Algebraic expressions consist of a combination of numbers, variables, and arithmetic operations. Understanding how to manipulate these expressions is crucial in solving algebraic problems. In the context of our task, expressions like \(3 - 5x + 4x^2\) require careful handling, especially when expanded further through operations like squaring \((a+h)\).
Expansion and simplification are key processes:

• Expansion: Breaking down expressions into simpler parts, like how \((a+h)^2\) expands into \(a^2 + 2ah + h^2\).
• Simplification: Combining like terms and reducing expressions to their simplest forms. After expansion and substitution, it's crucial to combine terms where possible.

These processes facilitate computations in finding the difference quotient \(\frac{f(a+h) - f(a)}{h}\). Cancellation of terms after substitution assists in reducing the complexity of the expression, making the end result easier to understand. Simplifying algebraic expressions helps in deriving meaningful insights from polynomial functions, ultimately supporting problem-solving in mathematics.