I am still confused on why they used the fraction -3/2 and -1.Can you explain further?THank you! ### Question 8Show 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 TheoremTo 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)

Theorem : Let f be continuous function on [a,b]. Let x=f(a) and y=f(b). Then we have for every z…

Answered: Show that the equation f(x) = 2x3 + 3x…

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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Question

I am still confused on why they used the fraction -3/2 and -1.

Can you explain further?

THank you!

### 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)

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