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: 34

Answer

The graph of function $f$ is shown below. The solution to equation $f(x)=0$ is $x\approx-1.7693$.
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Work Step by Step

$$f(\theta)=\theta^3-2\theta+2$$ a) Domain: $f(\theta)$ is defined on $R$. - Continuity: For all $\theta=c\in R$, we always have $$\lim_{\theta\to c}f(\theta)=c^3-2c+2=f(c)$$ Therefore, $f(\theta)$ is continuous on $R$. - At $\theta=-2$: $f(-2)=(-2)^3-2\times(-2)+2=-8+4+2=-2\lt0$ - At $\theta=0$: $f(0)=0^3-2\times0+2=0-0+2=2\gt0$ So $f(\theta)$ has changed its sign as $\theta$ changes from $-2$ to $0$. Since $f$ is continuous on $R$, according to the Intermediate Value Theorem, there is a value of $\theta=c\in[-2,0]$ for which $f(c)=0$. In other words, $f$ has a zero between $-2$ and $0$. b) The graph of function $f(\theta)$ is shown below. As noted in the graph, $f$ passes the $x$-axis at point $(-1.7693, 0)$. So the solution to the equation $f(x)=0$ is $x\approx-1.7693$. c) Evaluate the exact answer: $$\Big(\sqrt{\frac{19}{27}}-1\Big)^{1/3}-\Big(\sqrt{\frac{19}{27}}+1\Big)^{1/3}\approx-1.769292$$ The exact value here is close to the value found in part b).
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