University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.5 - Continuity - Exercises - Page 96: 76

Answer

The equation has 3 solutions: $(-3.906, 0.067, 3.839)$.
1536177482

Work Step by Step

$$x^3-15x+1=0$$ To prove the equation has a solution using the Immediate Value Theorem, we need to find 2 values $x_1$ and $x_2$ and show that there is a change of sign in the function $f(x)=x^3-15x+1$ as $x$ goes from $x_1$ to $x_2$. - Take $x_1=0$: $f(0)=0^3-15\times0+1=1\gt0$ - Take $x_2=2$: $f(2)=2^3-15\times2+1=8-30+1=-21\lt0$ So there is a change of sign in $f(x)$ as $x$ goes from $0$ to $2$. - On $[0,2]$: $\lim_{x\to c}f(x)=\lim_{x\to c}(x^3-15x+1)=c^3-15c+1=f(c)$ So $f(x)$ is continuous on $[0,2]$. Therefore, according to the Intermediate Value Theorem, there must be a value $x=c\in[0,2]$ such that $f(c)=0$. In other words, the equation $x^3-15x+1=0$ has at least one solution in the interval $[0,2]$. The graph of the function $f(x)=x^3-15x+1$ is enclosed below. Looking at the graph, it turns out that the curve $f(x)$ crosses the line $y=0$ at $3$ points. The $x$-coordinates of these points are the solutions of the equation. So the equation has 3 solutions: $(-3.906, 0.067, 3.839)$.
Small 1536177482
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.