Problem 35 - 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

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

Step 1: Find f(a)

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

Step 2: Find f(a+h)

Substitute \(a+h\) for \(x\) in the function to find \(f(a+h)\). This gives us \(f(a+h) = 3 - 5(a+h) + 4(a+h)^2\). Expand and simplify the expression: 1. Expand \((a+h)^2\): \((a+h)^2 = a^2 + 2ah + h^2\).2. Substitute and simplify:\(f(a+h) = 3 - 5a - 5h + 4(a^2 + 2ah + h^2)\).3. Now combine terms:\(f(a+h) = 3 - 5a - 5h + 4a^2 + 8ah + 4h^2\).

Step 3: Plug Values into the Difference Quotient

The difference quotient is given by \(\frac{f(a+h) - f(a)}{h}\). Plug in the expressions for \(f(a+h)\) and \(f(a)\) from the previous steps:\[ f(a+h) - f(a) = (3 - 5a - 5h + 4a^2 + 8ah + 4h^2) - (3 - 5a + 4a^2) \].Combine like terms:- The constant and \(a\) terms cancel: 3-3, 5a-5a, and 4a^2-4a^2.- Simplify the remaining terms:\(f(a+h) - f(a) = -5h + 8ah + 4h^2\).Finally, divide by \(h\):\( \frac{f(a+h) - f(a)}{h} = \frac{-5h + 8ah + 4h^2}{h} = -5 + 8a + 4h \).

Step 4: Simplify the Expression

The difference quotient simplifies to \(-5 + 8a + 4h\). Each term that was originally multiplied by \(h\) is now replaced by its constant factor, once the division by \(h\) occurs.

Key Concepts

Polynomial Functions

A polynomial function is a mathematical expression made up of terms that include variables raised to whole number powers and coefficients. In our exercise, the function given is \( f(x) = 3 - 5x + 4x^2 \). Here, we have:

• The constant term: 3
• A linear term: \(-5x\)
• A quadratic term: \(4x^2\)

Polynomial functions can have various degrees, which is determined by the highest power of the variable. The degree of this polynomial is 2 because the highest power of \(x\) is 2. Polynomial functions are fundamental in algebra because they can model a wide range of real-world scenarios, like kinetic energy or population growth. They are easy to differentiate and integrate, which makes them important for calculus too.

Function Evaluation

Function evaluation is the process of substituting a given value or expression into a function in place of the variable. In the exercise, we're asked to evaluate \(f(a)\) and \(f(a+h)\), starting with the polynomial function \(f(x) = 3 - 5x + 4x^2\).

• To find \(f(a)\), replace every \(x\) in \(f(x)\) with \(a\), giving us \(f(a) = 3 - 5a + 4a^2\).
• For \(f(a+h)\), substitute \(a+h\) into the function wherever \(x\) appears, and simplify:

1. Start by calculating \((a+h)^2\) which is \(a^2 + 2ah + h^2\).
2. Substitute \(a+h\) into the polynomial: \(f(a+h) = 3 - 5(a+h) + 4(a+h)^2\).
3. Simplify further to get \(f(a+h) = 3 - 5a - 5h + 4a^2 + 8ah + 4h^2\). Function evaluation helps us find the corresponding output of a function for specific inputs, which is a central concept in algebra.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying mathematical expressions to solve equations or evaluate functions. This skill includes adding, subtracting, multiplying, dividing, and expanding expressions.In this exercise, after finding \(f(a+h)\) and \(f(a)\), algebraic manipulation helps us compute the difference quotient:- The difference quotient in calculus is \(\frac{f(a+h) - f(a)}{h}\). It's a way to measure how a function changes as \(x\) changes, making it the backbone of the concept of the derivative.Here's how it's simplified:1. Subtract \(f(a) = 3 - 5a + 4a^2\) from \(f(a+h) = 3 - 5a - 5h + 4a^2 + 8ah + 4h^2\). Most terms cancel out, leaving \(-5h + 8ah + 4h^2\).2. Divide by \(h\) (since \(h eq 0\)) to simplify the expression to: \(-5 + 8a + 4h\).Algebraic manipulation is an essential skill for working with functions, making it important for both algebra and calculus fields.