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Find and simplify (a) \(f(x+h)-f(x)\) (b) \(\frac{f(x+h)-f(x)}{h}\) $$ f(x)=2-x^{2} $$

Short Answer

Expert verified
(a) \(f(x + h) - f(x) = - 2xh - h^2\) (b) \(\frac{f(x + h) - f(x)}{h} = - 2x - h\)

Step by step solution

01

Find f(x + h)

First, find \(f(x + h)\) by substituting \((x + h)\) into the function \(f(x) = 2 - x^2\): \[f(x + h) = 2 - (x + h)^2\]
02

Simplify f(x + h)

Now, we'll simplify \(f(x + h)\) by expanding the square: \[f(x + h) = 2 - (x^2 + 2xh + h^2)\] \[f(x + h) = 2 - x^2 - 2xh - h^2\]
03

Calculate f(x + h) - f(x)

Next, we'll subtract \(f(x)\) from \(f(x + h)\): \[(f(x + h) - f(x)) = (2 - x^2 - 2xh - h^2) - (2 - x^2)\]
04

Simplify f(x + h) - f(x)

Now, we'll simplify the expression for \(f(x + h) - f(x)\): \[(f(x + h) - f(x)) = - 2xh - h^2\]
05

Calculate \(\frac{f(x + h) - f(x)}{h}\)

Now, we'll move to part (b) and divide the expression for \((f(x + h) - f(x))\) by \(h\): \[\frac{f(x + h) - f(x)}{h} = \frac{- 2xh - h^2}{h}\]
06

Simplify \(\frac{f(x + h) - f(x)}{h}\)

Finally, we'll simplify the expression by canceling out a factor of \(h\): \[\frac{f(x + h) - f(x)}{h} = - 2x - h\] So, the simplified expressions are: (a) \(f(x + h) - f(x) = - 2xh - h^2\) (b) \(\frac{f(x + h) - f(x)}{h} = - 2x - h\)

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Key Concepts

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

Simplifying Expressions
In the realm of algebra, \textbf{simplifying expressions} is akin to tidying up a cluttered room – one sorts through all the elements, groups like terms, and condenses them to create a neater, more understandable form. This process often involves combining or canceling out terms and adhering to the fundamental properties of algebra.

Consider the function given in the exercise, which is a quadratic equation: \( f(x) = 2 - x^2 \). When tasked with simplifying an expression involving this function combined with another term, such as \( h \), we aim to reduce the expression to a form that's easier to work with or interpret. As seen in the steps provided for \( f(x + h) \), after expanding the square and subtracting \( f(x) \), the resulting expression is streamlined by removing zero pairs and combining like terms, reaching the simplified result, which reveals a linear dependence on \( h \).

One can apply this concept of simplification to a multitude of mathematical scenarios, honing an ability to parse complex equations and extract essential information.
Function Notation
When addressing \textbf{function notation}, we're essentially learning the language of functions. It's comparable to a code, whereby a set of rules is ascribed to an input to produce an output. In the exercise, \( f(x) \) represents the function name and the output, while \( x \) symbolizes the input variable. This notation lays the groundwork for understanding and communicating what happens to variables within the bounds of a function.

For instance, when we evaluate \( f(x + h) \), the \( x + h \) term is inserted into the place of \( x \) within the function, showing the effect of adding an increment \( h \) to the input before applying the function's rule. Understanding this notation is paramount, as it allows us to manipulate functions in algebraic operations, such as finding the difference quotient, with ease and clarity.
Algebraic Manipulation
At its heart, \textbf{algebraic manipulation} involves the various techniques used to rearrange and solve equations or expressions, which are the building blocks of algebra. Key tools in this arsenal include expanding brackets, factoring, and canceling terms, all of which serve to transform complex expressions into simpler, more manageable ones.

In the solution steps for part (b) of the exercise, we begin by dividing the difference \( f(x + h) - f(x) \) by \( h \), a procedure often required when dealing with the difference quotient or derivatives. By canceling out a common factor of \( h \) from both terms in the numerator, we end up with a more simplified expression, demonstrating clear-cut algebraic manipulation. Grasping these techniques is crucial, as they enable students to tackle a wide range of algebraic challenges with confidence and facilitate a deeper comprehension of calculus concepts such as derivatives and limits.

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