University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Practice Exercises - Page 111: 33


The graph of function $f$ is shown below. The solution to equation $f(x)=0$ is $x\approx1.3247$.

Work Step by Step

$$f(x)=x^3-x-1$$ a) Domain: $f(x)$ is defined on $R$. - Continuity: For all $x=c\in R$, we always have $$\lim_{x\to c}f(x)=c^3-c-1=f(c)$$ Therefore, $f(x)$ is continuous on $R$. - At $x=-1$: $f(-1)=(-1)^3-(-1)-1=-1+1-1=-1\lt0$ - At $x=2$: $f(2)=2^3-2-1=8-2-1=5\gt0$ So $f(x)$ has changed its sign as $x$ changes from $-1$ to $2$. Since $f$ is continuous on $R$, according to the Intermediate Value Theorem, there is a value of $x=c\in[-1,2]$ for which $f(c)=0$. In other words, $f$ has a zero between $-1$ and $2$. b) The graph of function $f(x)$ is shown below. As noted in the graph, $f$ passes the $x$-axis at point $(1.3247, 0)$. So the solution to the equation $f(x)=0$ is $x\approx1.3247$. c) Evaluate the exact answer: $$\Big(\frac{1}{2}+\frac{\sqrt{69}}{18}\Big)^{1/3}+\Big(\frac{1}{2}-\frac{\sqrt{69}}{18}\Big)^{1/3}\approx1.324718$$ The exact value here is close to the value found in part b).
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