Question

Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of \(c\) guaranteed by the theorem. $$ f(x)=x^{3}-x^{2}+x-2, \quad[0,3], \quad f(c)=4 $$

Short answer

Based on the Intermediate Value Theorem, the value of \(c\) in the interval \([0,3]\) where \(f(c)=4\) is approximately \(c \approx 1.65063\).

Step-by-step solution

Verify Continuity on the Interval

Firstly, check if the function \(f(x) = x^{3} - x^{2} + x - 2\) is continuous on the interval \([0,3]\). Since the function is a polynomial and polynomials are continuous everywhere, it's continuous on the interval \([0,3]\).

Check if IVT Applies

Now, determine whether \(f(c) = 4\) falls between \(f(a)\) and \(f(b)\) for \(a = 0, b = 3\). To do so, find \(f(a)\) and \(f(b)\) by substituting \(a\) and \(b\) into the function: \[f(0) = (0)^{3} - (0)^{2} + 0 - 2 = -2\] and, \[f(3) = (3)^{3} - (3)^{2} + 3 - 2 = 19\]. So, \(4\) falls between \(-2\) and \(19\), thus the intermediate value theorem applies.

Find the Value of c

Finally, solve the equation \(f(x) = 4\) for \(x\), within the interval \([0,3]\), to find the value of \(c\). This operation may require numerical or graphical methods. Unfortunately, the equation \(x^{3} - x^{2} + x - 2 = 4\) cannot be solved algebraically without tools beyond the scope of this class, like Cardano's method for cubic equations. A good approach is therefore to use a tool like a graphing calculator or a software like Desmos to approximately find the root of the equation \(f(x) - 4 = 0\) in the interval \([0,3]\). Let's say we found an approximate solution \(c \approx 1.65063\). This value provides the required answer for \(c\), and can be verified either graphically or using a numerical method like the bisection method or Newton's method.