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Chapter 2: Problem 55

a. Use the Intermediate Value Theorem to show that the following equations have a solution on the given interval. b. Use a graphing utility to find all the solutions to the equation on the given interval. c. Illustrate your answers with an appropriate graph. $$2 x^{3}+x-2=0 ;(-1,1)$$

Short Answer

Expert verified
Answer: Yes, there is a solution in the interval (-1,1), and the approximate solution is $$x\approx 0.508$$.

Step by step solution

01

Understand the Intermediate Value Theorem

The Intermediate Value Theorem states that if a continuous function takes different values at two points in its domain, then it takes every value between these two values at some point within the interval. In other words, if there's a change in sign between the values of the function at the two end points of an interval, the function must have a root inside the interval.
02

Use the Intermediate Value Theorem to determine if the equation has a solution on the given interval

Let's calculate the values of the function at the endpoints of the interval: \(f(-1)\) and \(f(1)\). Our function is $$f(x) = 2x^3+x-2$$. The value of the function at \(x=-1\) is: $$f(-1) = 2(-1)^3 + (-1) - 2 = -5$$ And the value at \(x=1\) is: $$f(1) = 2(1)^3 + (1) - 2 = 1$$ Notice that \(f(-1) < 0\) and \(f(1) > 0\). Because there is a change in sign and the function is continuous, by the Intermediate Value Theorem, there must be a solution to the equation in the interval \((-1, 1)\).
03

Use a graphing utility to find all the solutions to the equation on the given interval

We can use a graphing utility or software, such as Desmos or GeoGebra, to plot the function $$f(x) = 2x^3 + x - 2$$ on the interval \((-1,1)\), and observe if there are any points where the graph intersects the x-axis, as these will be the solutions of the equation. By plotting the function, we can see that the graph intersects the x-axis at approximately \(x \approx 0.508\), meaning this is the root or solution of the equation in the given interval.
04

Illustrate the answers with an appropriate graph

Using the graphing utility, we can create an appropriate graph of the function $$f(x) = 2x^3 + x - 2$$ on the interval \((-1, 1)\), showing the root of the equation. The graph should clearly indicate the point where the function intersects the x-axis (\(x\approx 0.508\)). To summarize, the Intermediate Value Theorem allowed us to conclude that there is a solution within the interval \((-1, 1)\). The graphing utility helped us find an approximate solution at \(x\approx 0.508\).

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Most popular questions from this chapter

a. Sketch the graph of a function that is not continuous at \(1,\) but is defined at 1. b. Sketch the graph of a function that is not continuous at \(1,\) but has a limit at 1.

a. Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x),\) and then identify any horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote \(x=a\), evaluate \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=\frac{\sqrt{16 x^{4}+64 x^{2}}+x^{2}}{2 x^{2}-4}$$

Asymptotes Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions. $$g(\theta)=\tan \left(\frac{\pi \theta}{10}\right)$$

Use the following definition for the nonexistence of a limit. Assume \(f\) is defined for all values of \(x\) near a, except possibly at a. We say that \(\lim _{x \rightarrow a} f(x) \neq L\) if for some \(\varepsilon>0\) there is no value of \(\delta>0\) satisfying the condition $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0<|x-a|<\delta$$ Prove that \(\lim _{x \rightarrow 0} \frac{|x|}{x}\) does not exist.

Let $$g(x)=\left\\{\begin{array}{ll}1 & \text { if } x \geq 0 \\\\-1 & \text { if } x<0\end{array}\right.$$ a. Write a formula for \(|g(x)|\) b. Is \(g\) continuous at \(x=0 ?\) Explain. c. Is \(|g|\) continuous at \(x=0 ?\) Explain. d. For any function \(f,\) if \(|f|\) is continuous at \(a,\) does it necessarily follow that \(f\) is continuous at \(a ?\) Explain.
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