Question

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

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

Step-by-step solution

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 \([0,3]\).

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 \(f(3) = (3)^{2} - 6*(3) + 8 = -1.\)

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.

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.