/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Find \(f(a+h)-f(a)\) for each fu... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(f(a+h)-f(a)\) for each function. Simplify your answer. \(f(x)=-\frac{1}{2} x+3\)

Short Answer

Expert verified
The simplified expression for \(f(a+h)-f(a)\) is \(-\frac{1}{2}h\).

Step by step solution

01

Find the expression for \(f(a)\)

We are given the function \(f(x) = -\frac{1}{2}x + 3\). To find \(f(a)\), we simply replace \(x\) with \(a\) in the function. So, \(f(a) = -\frac{1}{2}a + 3\).
02

Find the expression for \(f(a + h)\)

Now, we need to find the expression for \(f(a + h)\). We do this by replacing \(x\) with \((a + h)\) in the function. So, \(f(a+h) = -\frac{1}{2}(a+h) + 3\).
03

Calculate \(f(a + h) - f(a)\)

Next, we need to find the difference \(f(a + h) - f(a)\) and simplify the expression. \(f(a + h) - f(a) = \left(-\frac{1}{2}(a+h) + 3\right) - \left(-\frac{1}{2}a + 3\right)\)
04

Simplify the expression

Expand and simplify the expression: \(f(a + h) - f(a) = -\frac{1}{2}a - \frac{1}{2}h + 3 +\frac{1}{2}a - 3\) Notice that the \(-\frac{1}{2}a\) and the \(\frac{1}{2}a\) cancel each other, as do the \(+3\) and the \(-3\) terms. We are left with: \(f(a + h) - f(a) = -\frac{1}{2}h\) So, the final simplified expression for \(f(a + h) - f(a)\) is \(-\frac{1}{2}h\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
The process of function evaluation is about substituting specific values into a given function. It’s a fundamental concept in mathematics! For instance, if we have the function \( f(x) = -\frac{1}{2}x + 3 \), evaluating \( f(a) \) means that we substitute \( x \) with \( a \). This gives us the expression \( f(a) = -\frac{1}{2}a + 3 \).
- Understand that every function is like a machine, where you put a number in and get a number out.
- Substitute the variable (like \( x \)) with the given number (like \( a \)).
- Simplification might follow after substitution, often making the expression easier to handle.
Function evaluation is a core tool in calculus and algebra, allowing us to understand how functions behave under various circumstances. It becomes particularly useful when examining changes or differences in values, such as using \( f(a + h) \) in our example.
Simplification
Simplification involves making an expression easier to work with by reducing it to its most concise form. After evaluating a function, expressions often need to be simplified, especially in calculus.
For example, consider the difference \( f(a+h) - f(a) \). Initially, it might look a bit complicated. However, by simplifying we aim to remove any unnecessary complexity.
- Combine like terms. In our example, terms like \(-\frac{1}{2}a\) and \(\frac{1}{2}a\) cancel each other out.
- Discard constants when they sum to zero, such as \( +3 \) and \( -3 \) in the given problem.
- Focus on reducing terms so that the expression becomes clearer, which in our example is \(-\frac{1}{2}h\).
In algebra, simplification is often about transforming expressions into a form that is easier to understand and work with. This process can reveal insights about the problem and solutions that were not initially obvious.
Algebraic Expressions
An algebraic expression is a mathematical phrase that can contain numbers, variables, and operations (like addition, subtraction, etc.). Understanding how to manipulate these expressions is crucial.
In the exercise, you are given the function \( f(x) = -\frac{1}{2}x + 3 \), which is an algebraic expression. Here’s what makes up this expression and what you should know:
- **Variables**: Symbols like \( x \) or \( a \) that can change and represent unknown values.
- **Coefficients**: Numbers like \(-\frac{1}{2}\) that are multiplied by the variables.
- **Constants**: Fixed values, such as \(3\), not associated with any variables.
When dealing with algebraic expressions, it’s important to apply rules of arithmetic properly. This means distributing coefficients over terms properly, combining like terms, and simplifying wherever possible.
A strong grasp on handling algebraic expressions allows you to tackle more advanced problems with confidence, providing the foundational skills necessary for functions, calculus, and beyond.

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