Show that the equation f(x) = 2x3 + 3x + 6 = 0 has at least one solution; find an interval of length which must contain a solution.
Show that the equation f(x) = 2x3 + 3x + 6 = 0 has at least one solution; find an interval of length which must contain a solution.
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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I am still confused on why they used the fraction -3/2 and -1.
Can you explain further?
THank you!
Transcribed Image Text:### Question 8
Show that the equation \( f(x) = 2x^3 + 3x + 6 = 0 \) has at least one solution; find an interval of length \(\frac{1}{2}\) which must contain a solution.
![### Finding Roots Using the Intermediate Value Theorem
To determine where the function \( f(x) \) has roots, let’s look at the calculations step by step.
Consider the function \( f(x) = 2x^3 + 3x + 6 \).
1. Evaluate \( f(x) \) at \( x = -1 \):
\[ f(-1) = 2(-1)^3 + 3(-1) + 6 \]
\[ f(-1) = 2(-1) + (-3) + 6 \]
\[ f(-1) = -2 - 3 + 6 \]
\[ f(-1) = 1 \]
2. Evaluate \( f(x) \) at \( x = -\frac{3}{2} \):
\[ f\left( -\frac{3}{2} \right) = 2 \left( -\frac{3}{2} \right)^3 + 3 \left( -\frac{3}{2} \right) + 6 \]
\[ f\left( -\frac{3}{2} \right) = 2 \left( -\frac{27}{8} \right) + 3 \left( -\frac{3}{2} \right) + 6 \]
\[ f\left( -\frac{3}{2} \right) = 2 \left( -\frac{27}{8} \right) + \left( -\frac{9}{2} \right) + 6 \]
\[ f\left( -\frac{3}{2} \right) = -\frac{54}{8} - \frac{36}{8} + \frac{48}{8} \]
\[ f\left( -\frac{3}{2} \right) = - \frac{39}{4} \]
Since,
\[ f(-1) \cdot f\left( -\frac{3}{2} \right) < 0 \]
Hence, there exists some \( c \) in the interval \( (-\frac{3}{2}, -1) \) such that \( f(c) = 0 \).
This implies that \( f(x) \) has at least one root in the interval \( \left( -\frac{3}{2}, -1 \right)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc438c34-6e26-4f6a-9067-f367a43a4982%2F981ba14f-cf0f-47d7-aff9-d9c97b0dbb4a%2Fomqqloh.png&w=3840&q=75)
Transcribed Image Text:### Finding Roots Using the Intermediate Value Theorem
To determine where the function \( f(x) \) has roots, let’s look at the calculations step by step.
Consider the function \( f(x) = 2x^3 + 3x + 6 \).
1. Evaluate \( f(x) \) at \( x = -1 \):
\[ f(-1) = 2(-1)^3 + 3(-1) + 6 \]
\[ f(-1) = 2(-1) + (-3) + 6 \]
\[ f(-1) = -2 - 3 + 6 \]
\[ f(-1) = 1 \]
2. Evaluate \( f(x) \) at \( x = -\frac{3}{2} \):
\[ f\left( -\frac{3}{2} \right) = 2 \left( -\frac{3}{2} \right)^3 + 3 \left( -\frac{3}{2} \right) + 6 \]
\[ f\left( -\frac{3}{2} \right) = 2 \left( -\frac{27}{8} \right) + 3 \left( -\frac{3}{2} \right) + 6 \]
\[ f\left( -\frac{3}{2} \right) = 2 \left( -\frac{27}{8} \right) + \left( -\frac{9}{2} \right) + 6 \]
\[ f\left( -\frac{3}{2} \right) = -\frac{54}{8} - \frac{36}{8} + \frac{48}{8} \]
\[ f\left( -\frac{3}{2} \right) = - \frac{39}{4} \]
Since,
\[ f(-1) \cdot f\left( -\frac{3}{2} \right) < 0 \]
Hence, there exists some \( c \) in the interval \( (-\frac{3}{2}, -1) \) such that \( f(c) = 0 \).
This implies that \( f(x) \) has at least one root in the interval \( \left( -\frac{3}{2}, -1 \right)
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