Question

Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of \(c\) guaranteed by the theorem. $$ f(x)=\frac{x^{2}+x}{x-1}, \quad\left[\frac{5}{2}, 4\right], \quad f(c)=6 $$

Short answer

The IVT applies to the function \(f(x) = \frac{x^{2} + x}{x - 1}\) on the interval \([\frac{5}{2}, 4]\) and the value of \(c\) for which \(f(c) = 6\) is \(c = 4\).

Step-by-step solution

Verify Continuity on the Interval

Check the function \(f(x) = \frac{x^{2} + x}{x - 1}\) for continuity in the interval \([\frac{5}{2}, 4]\). It is observed that the function is not continuous at the point where the denominator equates to zero, i.e., \(x = 1\). Since \(1\) is not in the interval \([\frac{5}{2}, 4]\), it can be concluded that the function \(f(x)\) is continuous on the interval. The function is also continuous at the endpoints of the interval as the denominator is not zero at these points.

Evaluate Function at the Endpoints

Evaluate \(f(x)\) at \(\frac{5}{2}\) and \(4\). As per the definition of IVT, the value \(f(c) = 6\) must lie between \(f(\frac{5}{2})\) and \(f(4)\). Substituting \(\frac{5}{2}\) into the function, we get \(f(\frac{5}{2}) = \frac{\left(\frac{5}{2}\right)^{2}+\frac{5}{2}}{\frac{5}{2}-1} = 7\). Substituting \(4\) into function, we get \(f(4) = \frac{(4)^{2} + 4}{4 - 1} = 6\). It can be seen that \(f(c) = 6\) lies in between \(f(\frac{5}{2})\) and \(f(4)\), i.e., \(6 \leq f(c) \leq 7\).

Find Value of \(c\)

As per IVT, there exists a number \(c\) in the interval \([\frac{5}{2}, 4]\) such that \(f(c) = 6\). Since we found that \(f(4) = 6\), the value of \(c\) that satisfies \(f(c) = 6\) is \(c = 4\).