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Chapter 2: Problem 37

Use the Intermediate Value Theorem to show that each polynomial has a real zero between the given integers. $$f(x)=x^{3}+x^{2}-2 x+1 ; \text { between }-3 \text { and }-2$$

Short Answer

Expert verified
By applying the Intermediate Value Theorem, it is inferred that the function \(f(x)=x^{3}+x^{2}-2 x+1\) has a real zero between -3 and -2.

Step by step solution

01

Calculate the function value at the given points

Plug -3 and -2 into the function \(f(x)=x^{3}+x^{2}-2 x+1\) to get \(f(-3)\) and \(f(-2)\).\nA calculation for \(f(-3)\) can look like this: \[f(-3) = (-3)^3 + (-3)^2 - 2*(-3) + 1 = -27 + 9 + 6 + 1 = -11\]\nA similar calculation for \(f(-2)\) yields: \[f(-2) = (-2)^3 + (-2)^2 - 2*(-2) + 1 = -8 + 4 + 4 + 1 = 1\]
02

Analyzing the signs of the function values

By observing, \(f(-3) < 0\) and \(f(-2) > 0\), which means the function crosses the x-axis between -3 and -2.
03

Apply the Intermediate Value Theorem

Since \(f(-3)<0\) and \(f(-2)>0\), and since a polynomial is a continuous function, we apply the Intermediate Value Theorem which indicates there must be a root of the equation, or zero value for the function, in the interval (-3, -2).

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Most popular questions from this chapter

Use long division to rewrite the equation for \(g\) in the form $$\text {quotient }+\frac{\text {remainder}}{\text {divisor}}$$ Then use this form of the function's equation and transformations of \(f(x)=\frac{1}{x}\) to graph \(g.\) $$g(x)=\frac{3 x+7}{x+2}$$

Use transformations of \(f(x)=\frac{1}{x}\) or \(f(x)=\frac{1}{x^{2}}\) to graph each rational function. $$g(x)=\frac{1}{x+1}-2$$

Use transformations of \(f(x)=\frac{1}{x}\) or \(f(x)=\frac{1}{x^{2}}\) to graph each rational function. $$h(x)=\frac{1}{x}+2$$

Find the horizontal asymptote, if there is one, of the graph of rational function. $$f(x)=\frac{-2 x+1}{3 x+5}$$

The equation for \(f\) is given by the simplified expression that results after performing the indicated operation. Write the equation for \(f\) and then graph the function. $$\frac{1-\frac{3}{x+2}}{1+\frac{1}{x-2}}$$
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