/*! 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 79 \(f(x)=x^{2}-x+1, \quad \frac{f(... [FREE SOLUTION] | 91Ó°ÊÓ

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\(f(x)=x^{2}-x+1, \quad \frac{f(2+h)-f(2)}{h}, h \neq 0\)

Short Answer

Expert verified
The solution to \(\frac{f(2+h)-f(2)}{h}\) is \( h + 3 \).

Step by step solution

01

Calculate \(f(2+h)\)

First, substitute \(x=2+h\) into the function \(f(x)=x^{2}-x+1\). This gives us: \(f(2+h) = (2+h)^2 - (2+h) + 1 = 4 + 4h + h^2 - 2 - h + 1 = h^2 + 3h + 3\).
02

Calculate \(f(2)\)

Next, substitute \(x=2\) into the function \(f(x)=x^{2}-x+1\). This gives us: \(f(2) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3\).
03

Substitute \(f(2+h)\) and \(f(2)\) into the difference quotient

Substitute the calculated values into the difference quotient formula given by \(\frac{f(2+h)-f(2)}{h}\), resulting in \( \frac{(h^2 + 3h + 3) - 3}{h} = h + 3\).

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Key Concepts

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

Polynomial Functions
Polynomial functions form the foundation of many algebraic problems in mathematics. They consist of variables raised to whole number exponents and coefficients that are typically real numbers. A general polynomial function can be expressed in the form:
  • \[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \ \,\ldots\, \,+ a_1x + a_0 \ \\]
  • Here, \(a_n, a_{n-1},\) ..., \(a_0\) are coefficients and \(n\) is a non-negative integer, indicating the degree of the polynomial.
In the given exercise, the polynomial function is \(f(x) = x^2 - x + 1\). This is a quadratic polynomial function because the highest power of \(x\) is 2.

This means the function describes a parabolic curve when graphed on a coordinate plane. Polynomial functions like the one in this exercise are smooth and continuous. They are often used to model real-world phenomena, and understanding their behavior is crucial for solving complex equations.
Derivatives
Derivatives are a key concept in calculus that describe the rate of change of a function. Simply put, a derivative shows how a function changes as its input changes. For polynomial functions, the derivative can be calculated using rules that simplify the process.
  • The power rule is often used to find derivatives of polynomial functions: \[ \frac{d}{dx} x^n = nx^{n-1} \]
In the context of the exercise, the calculation of the difference quotient is an essential step towards finding the derivative of the function.

When we compute the difference quotient, \(\frac{f(2+h) - f(2)}{h}\), where \(h eq 0\), we are effectively determining the slope of the secant line over the interval between \(x = 2\) and \(x = 2 + h\).
This slope provides an approximation of the derivative at \(x = 2\), which reflects how the function is behaving at that specific point.
Limit Process
The concept of a limit is a cornerstone in calculus and is essential in the definition of derivatives. The limit process helps in transitioning from the difference quotient to the derivative.
  • In the formula \(\frac{f(2+h) - f(2)}{h}\), as \(h\) approaches zero, the value of the difference quotient approaches the actual slope of the tangent line at \(x = 2\).
  • This approach is formally expressed as:\[ \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} \]
This limit gives us the derivative \(f'(2)\), providing the exact rate of change of the function at \(x = 2\).
Understanding limits is crucial because they allow us to analyze and understand the behavior of functions at specific points, especially where the function changes direction or has a stationary point. The limit process bridges the gap between algebraic calculations and the intuitive concept of instantaneous rate of change.

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Most popular questions from this chapter

Transportation For groups of 80 or more people, a charter bus company determines the rate per person according to the formula Rate \(=8-0.05(n-80), \quad n \geq 80\) where the rate is given in dollars and \(n\) is the number of people. (a) Write the revenue \(R\) for the bus company as a function of \(n\). (b) Use the function in part (a) to complete the table. What can you conclude? \begin{tabular}{|l|l|l|l|l|l|l|l|} \hline\(n\) & 90 & 100 & 110 & 120 & 130 & 140 & 150 \\ \hline\(R(n)\) & & & & & & & \\ \hline \end{tabular}

In Exercises 55-68, determine whether the function has an inverse function. If it does, find the inverse function. $$ p(x)=-4 $$

Cost, Revenue, and Profit A company produces a product for which the variable cost is \(\$ 12.30\) per unit and the fixed costs are \(\$ 98,000\). The product sells for \(\$ 17.98\). Let \(x\) be the number of units produced and sold. (a) The total cost for a business is the sum of the variable cost and the fixed costs. Write the total cost \(C\) as a function of the number of units produced. (b) Write the revenue \(R\) as a function of the number of units sold. (c) Write the profit \(P\) as a function of the number of units sold. (Note: \(P=R-C\) )

The total numbers \(f\) (in billions) of miles traveled by motor vehicles in the United States from 1995 through 2002 are shown in the table. The time (in years) is given by \(t\), with \(t=5\) corresponding to 1995 . (Source: U.S. Federal Highway Administration) $$ \begin{array}{|c|c|} \hline 0 \text { Year, } t & \text { Miles traveled, } f(t) \\ \hline 5 & 2423 \\ 6 & 2486 \\ 7 & 2562 \\ 8 & 2632 \\ 9 & 2691 \\ 10 & 2747 \\ 11 & 2797 \\ 12 & 2856 \\ \hline \end{array} $$ (a) Does \(f^{-1}\) exist? (b) If \(f^{-1}\) exists, what does it mean in the context of the problem? (c) If \(f^{-1}\) exists, find \(f^{-1}\) (2632). (d) If the table was extended to 2003 and if the total number of miles traveled by motor vehicles for that year was 2747 billion, would \(f^{-1}\) exist? Explain.

In Exercises 39-54, (a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=\sqrt{4-x^{2}}, \quad 0 \leq x \leq 2 $$

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