Question

In Exercises 65 and 66, apply the Extended Mean Value Theorem to the functions \(f\) and \(g\) on the given interval. Find all values \(c\) in the interval \((a, b)\) such that \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\) \(\begin{array}{l} \underline{\text { Functions }} \\ f(x)=\sin x, \quad g(x)=\cos x \end{array} \quad \frac{\text { Interval }}{\left[0, \frac{\pi}{2}\right]}\)

Short answer

The value of \(c\) that satisfies the given conditions of the Extended Mean Value Theorem for the given functions and interval is \(c = \frac{\pi}{4}\).

Step-by-step solution

Calculate the derivatives

First, the derivatives of the functions \(f(x) = \sin x\) and \(g(x) = \cos x\) are needed. The derivative of \(f(x) = \sin x\) is \(f^{\prime}(x) = \cos x\) and the derivative of \(g(x) = \cos x\) is \(g^{\prime}(x) = -\sin x\).

Calculate \(f(b) - f(a)\) and \(g(b) - g(a)\)

In the next step, calculate the differences in the values of \(f\) and \(g\) at \(b\) and \(a\) respectively. Here, \(f(b) - f(a) = f\(\frac{\pi}{2}\) - f(0)= 1 - 0 = 1\), and \(g(b) - g(a) = g\(\frac{\pi}{2}\) - g(0) = 0 - 1 = -1\).

Apply the Extended Mean Value Theorem

According to the EMVT, there exists a \(c\) in the interval \((0, \frac{\pi}{2})\) where \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\). Substituting the given values, we get: \(\frac{\cos c}{-\sin c} = \frac{1}{-1}\). Solving this, we get: \(\cos c = \sin c\).

Find \(c\) satisfying the equation.

The equation \(\sin c = \cos c\) is satisfied when \(c = \frac{\pi}{4}\). However, it's important to note other solutions for this equation exist, but they fall outside the interval \((0, \frac{\pi}{2})\). Therefore, \(c = \frac{\pi}{4}\) is the only solution within this interval.