Prove that f(x) = 5x? + 3x – 26 is 0(x²). %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Prove that \( f(x) = 5x^2 + 3x - 26 \) is \( \theta(x^2) \).
---
Follow the steps below to prove that the given function \( f(x) \) belongs to the asymptotic notation \( \theta(x^2) \).
1. **Definition of \( \theta \)-notation:**
A function \( f(x) \) is said to be \( \theta(g(x)) \) if there are positive constants \( c_1 \), \( c_2 \), and \( n_0 \) such that for all \( x \geq n_0 \),
\[
c_1 g(x) \leq f(x) \leq c_2 g(x)
\]
2. **Apply the definition to \( f(x) \):**
We need to show that there exist constants \( c_1 \), \( c_2 \), and \( n_0 \) such that
\[
c_1 x^2 \leq 5x^2 + 3x - 26 \leq c_2 x^2
\]
for all \( x \geq n_0 \).
3. **Find the upper bound:**
To find an upper bound, observe that the term \( 5x^2 \) will dominate for larger values of \( x \). Therefore,
\[
5x^2 + 3x - 26 \leq 5x^2 + 3x
\]
\[
5x^2 + 3x \leq 5x^2 + 3x \leq 6x^2 \quad \text{(for sufficiently large x)}
\]
For \( c_2 = 6 \), let's find \( n_0 \) where this is valid. For large \( x \), \( 3x \) becomes negligible compared to \( 5x^2 \).
4. **Find the lower bound:**
Similarly, for the lower bound:
\[
5x^2 + 3x - 26 \geq 5x^2 - 26
\]
\[
5x^2 - 26 \geq 4x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb7f7394f-2209-482a-a19e-9ec4a0e092ef%2F8219d27d-ee7d-4fa0-8d26-0694ab3c08c5%2Fucch0cq_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Prove that \( f(x) = 5x^2 + 3x - 26 \) is \( \theta(x^2) \).
---
Follow the steps below to prove that the given function \( f(x) \) belongs to the asymptotic notation \( \theta(x^2) \).
1. **Definition of \( \theta \)-notation:**
A function \( f(x) \) is said to be \( \theta(g(x)) \) if there are positive constants \( c_1 \), \( c_2 \), and \( n_0 \) such that for all \( x \geq n_0 \),
\[
c_1 g(x) \leq f(x) \leq c_2 g(x)
\]
2. **Apply the definition to \( f(x) \):**
We need to show that there exist constants \( c_1 \), \( c_2 \), and \( n_0 \) such that
\[
c_1 x^2 \leq 5x^2 + 3x - 26 \leq c_2 x^2
\]
for all \( x \geq n_0 \).
3. **Find the upper bound:**
To find an upper bound, observe that the term \( 5x^2 \) will dominate for larger values of \( x \). Therefore,
\[
5x^2 + 3x - 26 \leq 5x^2 + 3x
\]
\[
5x^2 + 3x \leq 5x^2 + 3x \leq 6x^2 \quad \text{(for sufficiently large x)}
\]
For \( c_2 = 6 \), let's find \( n_0 \) where this is valid. For large \( x \), \( 3x \) becomes negligible compared to \( 5x^2 \).
4. **Find the lower bound:**
Similarly, for the lower bound:
\[
5x^2 + 3x - 26 \geq 5x^2 - 26
\]
\[
5x^2 - 26 \geq 4x
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