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Chapter 1: Problem 27

$$f(x)=\frac{x^{2}-2 x-15}{x^{3}-5 x^{2}+x-5}$$

Short Answer

Expert verified
\(f(x)=\frac{(x-5)(x+3)}{(x-3)(x-2)(x+1)}\)

Step by step solution

01

Factor the numerator and denominator

Factor the numerator \(x^{2}-2 x-15\) to get \((x-5)(x+3)\). Then, factor the denominator \(x^{3}-5 x^{2}+x-5\) to get \((x-3)(x-2)(x+1)\). So the function becomes \(f(x)=\frac{(x-5)(x+3)}{(x-3)(x-2)(x+1)}\).
02

Cancel out common factors

There are no common factors in the numerator and denominator that can be cancelled out. So, the function remains the same.

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Key Concepts

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

Factoring Polynomials
Factoring polynomials is a fundamental skill you need to master in algebra. This involves breaking down a complex polynomial into simpler parts, called factors, that multiply together to give the original polynomial. For example, when you have a polynomial like 2-2x-15>, the aim is to express it as (x-a)(x-b), where a and b are numbers.
This process makes it easier to work with polynomials, especially in rational functions.
  • To factorize, look for two numbers that both add up to -2 (the coefficient of x) and multiply to give -15 (the constant term).
  • Here, those numbers are -5 and 3, resulting in the factors (x-5)(x+3).
Similarly, the denominator in the function is a cubic polynomial, 3-5x2+x-5>. Factoring the denominator means identifying multiple factors. In practice, you look for values that cancel out each other when expanded, reducing computation. These factors turn out to be (x-3)(x-2)(x+1).
This method of factoring allows us to simplify mathematical expressions and find roots more efficiently.
Numerator and Denominator
In a rational function, the numerator and denominator play distinct roles. The numerator of a rational function sits on the top in the fraction, whereas the denominator sits on the bottom. For the given problem, the numerator is 2-2x-15> and the denominator is 3-5x2+x-5>.
Understanding the behavior of each helps in solving and simplifying these functions.
  • The numerator defines the multiplicative part of the function. It influences the overall value of the function directly based on the value of x.
  • The denominator is critical because it determines the domain of the function鈥攖he set of possible values for x.
The denominator cannot be zero as it makes the function undefined. This can lead to restrictions in the values x can take, crucial for graphing the function and identifying potential asymptotes.
Simplifying Rational Expressions
Simplifying rational expressions entails reducing fractions to their simplest form. With polynomials, this means identifying any common factors in the numerator and the denominator, then cancelling them out. In the exercise, the initial function is\[ f(x)=\frac{(x-5)(x+3)}{(x-3)(x-2)(x+1)} \]Upon looking at the expression, we notice that there are no common factors between the numerator and the denominator in this example.
So, the expression is already at its simplest form.
  • Cancelling common factors is streamlined by factoring, which we explored in previous sections.
  • While simplifying, always check that no term in the denominator equals zero with given variable values.
Simplifying these expressions is key for making complex functions easier to analyze and understand. Plus, it highlights any existing domain restrictions where the denominator becomes zero.

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Most popular questions from this chapter

On a trip of \(d\) miles to another city, a truck driver's average speed was \(x\) miles per hour. On the return trip, the average speed was \(y\) miles per hour. The average speed for the round trip was 50 miles per hour. (a) Verify that $$\begin{array}{l}{y=\frac{25 x}{x-25}} \\ {\text { What is the domain? }}\end{array}$$ What is the domain? (b) Complete the table. Are the values of y different than you expected? Explain. (c) Find the limit of y as x approaches 25 from the right and interpret its meaning.

Using the Intermediate Value Theorem In Exercises 89-94, use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval [0, 1]. Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$f(x)=\sqrt{x^{4}+39 x+13}-4$$

Signum Function The signum function is defined by $$\operatorname{sgn}(x)=\left\\{\begin{array}{ll}{-1,} & {x<0} \\ {0,} & {x=0} \\ {1,} & {x>0}\end{array}\right.$$ Sketch a graph of \(\operatorname{sgn}(x)\) and find the following (if possible). $$(a) lim _{x \rightarrow 0} \operatorname{sgn}(x) \quad \text { (b) } \lim _{x \rightarrow 0^{+}} \operatorname{sgn}(x) \quad \text { (c) } \lim _{x \rightarrow 0} \operatorname{sgn}(x)$$

Free-Falling Object In Exercises 103 and \(104,\) use the position function \(s(t)=-4.9 t^{2}+200,\) which gives the height (in meters) of an object that has fallen for \(t\) seconds from a height of 200 meters. The velocity at time \(t=a\) seconds is given by $$\lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t}$$ Find the velocity of the obiect when \(t=3\)

A sporting goods manufacturer designs a golf ball having a : volume of 2.48 cubic inches. (a) What is the radius of the golf ball? (b) The volume of the golf ball varies between 2.45 cubic inches. How does the radius vary? (c) Use the \(\varepsilon-\delta\) definition of limit to describe this situation. Identify \(\varepsilon\) and \(\delta .\)
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