/*! 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 27 (a) Find the difference quotient... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Find the difference quotient \(\frac{f(x)-f(a)}{x-a}\) for each function, as in Example 4. (b) Find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for each function, as in Example \(5 .\) $$f(x)=x^{2}-2 x+4$$

Short Answer

Expert verified
(a) Difference quotient is \( x + a - 2 \). (b) Difference quotient is \( 2x + h - 2 \).

Step by step solution

01

Simplify f(x)

Given the function \( f(x) = x^2 - 2x + 4 \). The first task is to evaluate \( f(a) \), which is the same expression where \( x \) is replaced by \( a \). Therefore, \( f(a) = a^2 - 2a + 4 \).
02

Calculate Difference Quotient for (a)

The difference quotient for part (a) is \( \frac{f(x) - f(a)}{x-a} \).Substitute \( f(x) = x^2 - 2x + 4 \) and \( f(a) = a^2 - 2a + 4 \) into the difference quotient:\[ \frac{(x^2 - 2x + 4) - (a^2 - 2a + 4)}{x-a} \]Simplify the numerator:\[ x^2 - 2x + 4 - a^2 + 2a - 4 = (x^2 - a^2) - 2(x - a) \]Factor the numerator:\[ \frac{(x-a)(x+a) - 2(x-a)}{x-a} = \frac{(x-a)(x+a-2)}{x-a} \]Cancel \((x-a)\) in the numerator and denominator:\[ x+a-2 \]
03

Calculate Difference Quotient for (b)

For part (b), we find \( \frac{f(x+h)-f(x)}{h} \)Calculate \( f(x+h) \):\[ f(x+h) = (x+h)^2 - 2(x+h) + 4 = x^2 + 2xh + h^2 - 2x - 2h + 4 \]Substitute \( f(x) \) and \( f(x+h) \) into the difference quotient:\[ \frac{(x^2 + 2xh + h^2 - 2x - 2h + 4) - (x^2 - 2x + 4)}{h} \]Simplify the expression:\[ \frac{2xh + h^2 - 2h}{h} \]Factor \( h \) in the numerator:\[ \frac{h(2x + h - 2)}{h} \]Cancel \( h \) in the numerator and denominator:\[ 2x + h - 2 \]

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Key Concepts

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

Understanding Algebra Through Difference Quotients
Algebra is the foundation of many mathematical concepts, including difference quotients. The difference quotient is a method used in algebra to approximate derivatives, which helps us understand how functions change. It involves two main difference quotient formulas:
  • The first formula is \( \frac{f(x)-f(a)}{x-a} \).
  • The second formula is \( \frac{f(x+h)-f(x)}{h} \).
This allows us to see the change in the function value, \( f(x) \), over a small interval. By substituting values into these formulas, we simplify expressions and grasp the behavior of the function. In the original exercise, applying algebraic manipulation such as factoring and simplifying rational expressions is crucial to solve the difference quotients.
Function Evaluation: Plugging In Values
Function evaluation is all about substituting a variable in a function with a specific number or expression. This process helps us find the output or result when the input changes. With a function like \( f(x) = x^2 - 2x + 4 \), to evaluate \( f(a) \), we replace every \( x \) with \( a \). This gives us \( f(a) = a^2 - 2a + 4 \).
In the difference quotient exercise, evaluating \( f(a) \) and \( f(x+h) \) is essential. This process allows us to simplify the difference quotients by substituting these evaluated functions into the formula, offering insight into the rate at which the function changes as \( x \) or \( h \) varies.
Exploring Polynomials in Difference Quotients
Polynomials, such as \( f(x) = x^2 - 2x + 4 \), often appear in difference quotient problems. A polynomial is an expression consisting of variables and coefficients, structured in terms of powers of the variables. For the given polynomial, applying the difference quotient involves handling terms like \( x^2 \), \( x \), and constant terms like 4.
Using difference quotients, we can determine how a polynomial function's output changes as the input alters slightly. Simplifying expressions involving polynomials requires us to collect like terms and perform operations like expansion and factoring. Polynomials are key in making the difference quotient tangible by providing real examples to work with.
Navigating Rational Expressions
Rational expressions are fractions where the numerator and the denominator are both polynomials. When dealing with difference quotients, simplifying rational expressions is a core skill needed to manipulate and reduce fractions into a simpler form.
In the given problem, the expressions like \( \frac{(x-a)(x+a-2)}{x-a} \) and \( \frac{h(2x + h - 2)}{h} \) are rational expressions. To simplify these, we perform operations like factoring, canceling common factors, and reducing the expressions to reveal clear conclusions like \( x+a-2 \) or \( 2x+h-2 \). Understanding rational expressions helps in the accurate computation of difference quotients, making complex algebra problems more manageable.

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Most popular questions from this chapter

Let the function \(L\) be defined by the following rule: \(L(x)\) is the exponent to which 2 must be raised to yield \(x\). (For the moment, we won't concern ourselves with the domain and range.) Then \(L(8)=3,\) for example, since the exponent to which 2 must be raised to yield 8 is 3 (that is, \(8=2^{3}\) ). Find the following outputs (a) \(L(1)\) (b) \(L(2)\) (c) \(L(4)\) (d) \(L(64)\) (e) \(L(1 / 2)\) (f) \(L(1 / 4)\) (g) \(L(1 / 64)\) (h) \(L(\sqrt{2})\)

Let \(g(t)=|t-4| .\) Find \(g(3) .\) Find \(g(x+4)\)

Use the given function and compute the first six iterates of each initial input \(x_{0}\). In cases in which a calculator answer contains four or more decimal places, round the final answer to three decimal places. (However, during the calculations, work with all of the decimal places that your calculator affords.) \(f(x)=2 x\) (a) \(x_{0}=1\) (b) \(x_{0}=0\) (c) \(x_{0}=-1\)

Suppose that in a certain biology lab experiment, the number of bacteria is related to the temperature \(T\) of the environment by the function $$N(T)=-2 T^{2}+240 T-5400 \quad(40 \leq T \leq 90)$$ Here, \(N(T)\) represents the number of bacteria present when the temperature is \(T\) degrees Fahrenheit. Also, suppose that \(t\) he experiment begins, the temperature is given by $$T(t)=10 t+40 \quad(0 \leq t \leq 5)$$ (a) Compute \(N[T(t)]\) (b) How many bacteria are present when \(t=0\) hr? When \(t=2 \mathrm{hr} ?\) When \(t=5 \mathrm{hr} ?\)

Consider the following two rules, \(F\) and \(G,\) where \(F\) is the rule that assigns to each person his or her birth-mother and \(G\) is the rule that assigns to each person his or her aunt. Explain why \(F\) is a function but \(G\) is not.

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