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Chapter 1: Problem 66

Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of \(c\) guaranteed by the theorem. $$ f(x)=x^{2}-6 x+8, \quad[0,3], \quad f(c)=0 $$

Short Answer

Expert verified
The Intermediate Value Theorem applies to the interval, and the value of \(c\) that satisfies the theorem is \(2\).

Step by step solution

01

Check if Function is Continuous

Check whether the function \(f(x) = x^{2} - 6x + 8\) is continuous in the given interval \([0,3]\). As a polynomial function, it is continuous on the entire real line, including the interval \([0,3]\).
02

Calculate f(a) and f(b)

Next, calculate \(f(a)\) and \(f(b)\) where \(a=0\) and \(b=3\). \n For \(f(0)\), substitute \(0\) into the function to get \(f(0) = (0)^{2} - 6*(0) + 8 = 8\). \n For \(f(3)\), substitute \(3\) into the function to get \(f(3) = (3)^{2} - 6*(3) + 8 = -1.\)
03

Check if IVT Applies

Intermediate Value Theorem applies if there's a number \(k\) that lies between \(f(a)\) and \(f(b)\). Here, \(k=0\), and it lies between \(f(a)=8\) and \(f(b)=-1\). Therefore, the IVT applies here.
04

Solve for c

Finally, to find the \(c\) that satisfies IVT, solve the equation \(f(c)=0\). So, \(c^{2} - 6c + 8 = 0\). Factoring the quadratic equation results in \((c-2)(c-4)=0\). So the roots are \(c=2\) and \(c=4\). However, only \(c=2\) is in the interval \([0,3]\). Therefore, \(c=2\) is the solution.

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Most popular questions from this chapter

Explain why the function has a zero in the given interval. $$ \begin{array}{lll} \text { Function } & \text { Interval } \\ f(x)=x^{3}+3 x-2 & {[0,1]} \\ \end{array} $$

Use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval [0, 1]. Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$ f(x)=x^{3}+x-1 $$

Give an example of two functions that agree at all but one point.

Write the expression in algebraic form. \(\sin (\operatorname{arcsec} x)\)

Write the expression in algebraic form. \(\csc \left(\arctan \frac{x}{\sqrt{2}}\right)\)
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