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 95: 57

Answer

Use the Intermediate Value Theorem to show that there exists values of $c$ such that $f(c)$ equals the given values.
1536151498

Work Step by Step

$$f(x)=x^3-8x+10$$ - Domain: $(-\infty,\infty)$ 1) Continuity: On $(-\infty,\infty)$, we notice that $\lim_{x\to c}(x^3-8 x+10)=c^3-8c+10=f(c)$ That means $f$ is continuous on $(-\infty,\infty)$. 2) Show that $c$ exists which $f(c)=\pi$ Here I graph $f(x)=x^3-8 x+10$ for better findings, which I have also enclosed below. - At $x=0$: $f(0)=0^3-8(0)+10=10$ - At $x=2$: $f(2)=2^3-8(2)+10=8-16+10=2$ Here, since $2\lt\pi\lt10$ and $f$ is continuous on $[0,2]$, according to Intermediate Value Theorem, there exists a value of $x=c\in[0,2]$ such that $f(c)=\pi$. 3) Show that $c$ exists which $f(c)=-\sqrt3$ - At $x=0$: $f(0)=0^3-8(0)+10=10$ - At $x=-4$: $f(-4)=(-4)^3-8(-4)+10=-64+32+10=-22$ Here, since $-22\lt-\sqrt3\lt10$ and $f$ is continuous on $[-4,0]$, according to Intermediate Value Theorem, there exists a value of $x=c\in[-4,0]$ such that $f(c)=-\sqrt3$. 4) Show that $c$ exists which $f(c)=5,000,000$ - At $x=0$: $f(0)=0^3-8(0)+10=10$ - At $x=1000$: $f(1000)=1000^3-8(1000)+10=999,992,010$ Here, since $10\lt5,000,000\lt999,992,010$ and $f$ is continuous on $[0,1000]$, according to Intermediate Value Theorem, there exists a value of $x=c\in[0,1000]$ such that $f(c)=5,000,000$.
Small 1536151498
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.