State the domain and range of the function. x-8x+15 A f(x) = X-3 The domain of the function is. (Type your answer in interval notation.) The range of the function is (Type your answer in interval notation.)
State the domain and range of the function. x-8x+15 A f(x) = X-3 The domain of the function is. (Type your answer in interval notation.) The range of the function is (Type your answer in interval notation.)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
![### Determining the Domain and Range of a Function
**Problem:**
State the domain and range of the function.
\[ f(x) = \frac{x^2 - 8x + 15}{x - 3} \]
**Solution:**
1. **Finding the Domain:**
The domain of the function includes all real numbers except where the denominator is zero. For \( f(x) = \frac{x^2 - 8x + 15}{x - 3} \), the denominator is zero when \( x = 3 \).
Therefore, the domain of the function is all real numbers except \( x = 3 \).
**Domain in Interval Notation:**
\[ (-\infty, 3) \cup (3, \infty) \]
2. **Finding the Range:**
To find the range, we need to analyze the behavior of the function and any values that \( f(x) \) cannot take. This often involves simplifying the expression and determining any horizontal asymptotes or excluded values.
When we simplify the expression:
\[ f(x) = \frac{(x - 3)(x - 5)}{x - 3} \]
For \( x \ne 3 \),
\[ f(x) = x - 5 \]
The simplified function, \( f(x) = x - 5 \), is a linear function with a slope of 1 and a y-intercept of -5, defined for all \( x \) except for \( x = 3 \).
As \( x \to 3 \), the original function approaches:
\[ f(3) = \lim_{x \to 3} (x - 5) \]
Because the original form has a discontinuity at \( x = 3 \), we need to confirm whether \( f(x) \) can reach every real number or if there are exceptions.
Given that \( x - 5 \) can take any real number value, and there is no vertical asymptote impacting the range, the range is all real numbers.
**Range in Interval Notation:**
\[ (-\infty, \infty) \]
**Results:**
- **The domain of the function is \( (-\infty, 3) \cup (3, \infty) \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F573db276-7549-444e-9b95-52b12d1d154b%2Fdbc5370c-7c59-4b89-a7da-daf55027a1b4%2F5zagyre_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining the Domain and Range of a Function
**Problem:**
State the domain and range of the function.
\[ f(x) = \frac{x^2 - 8x + 15}{x - 3} \]
**Solution:**
1. **Finding the Domain:**
The domain of the function includes all real numbers except where the denominator is zero. For \( f(x) = \frac{x^2 - 8x + 15}{x - 3} \), the denominator is zero when \( x = 3 \).
Therefore, the domain of the function is all real numbers except \( x = 3 \).
**Domain in Interval Notation:**
\[ (-\infty, 3) \cup (3, \infty) \]
2. **Finding the Range:**
To find the range, we need to analyze the behavior of the function and any values that \( f(x) \) cannot take. This often involves simplifying the expression and determining any horizontal asymptotes or excluded values.
When we simplify the expression:
\[ f(x) = \frac{(x - 3)(x - 5)}{x - 3} \]
For \( x \ne 3 \),
\[ f(x) = x - 5 \]
The simplified function, \( f(x) = x - 5 \), is a linear function with a slope of 1 and a y-intercept of -5, defined for all \( x \) except for \( x = 3 \).
As \( x \to 3 \), the original function approaches:
\[ f(3) = \lim_{x \to 3} (x - 5) \]
Because the original form has a discontinuity at \( x = 3 \), we need to confirm whether \( f(x) \) can reach every real number or if there are exceptions.
Given that \( x - 5 \) can take any real number value, and there is no vertical asymptote impacting the range, the range is all real numbers.
**Range in Interval Notation:**
\[ (-\infty, \infty) \]
**Results:**
- **The domain of the function is \( (-\infty, 3) \cup (3, \infty) \)
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning