Problem 44 - Essential Calculus

Question

Let \(f\) be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). Also, suppose that \(f(a)=f(b)\) and that \(c\) is a real number in the interval such that \(f^{\prime}(c)=0 .\) Find an interval for the function \(g\) over which Rolle's Theorem can be applied, and find the corresponding critical number of \(g(k\) is a constant). (a) \(g(x)=f(x)+k\) (b) \(g(x)=f(x-k)\) (c) \(g(x)=f(k x)\)

Step-by-Step Solution

Answer

Rolle's Theorem can be directly applied to \(g(x) = f(x) + k\) over the interval \([a, b]\) with \(c\) being the corresponding critical number. For the other two functions, \(g(x) = f(x - k)\) and \(g(x) = f(kx)\), the theorem's applicability and the determination of corresponding critical numbers will depend upon the specific values of \(a, b, k\).

Step 1: Determining interval for \(g(x)=f(x)+k\)

For \(g(x) = f(x) + k\), we can apply Rolle's Theorem over the interval \([a, b]\). The constant \(k\) shifts the function vertically but doesn't affect the derivative, hence \(g'(x) = f'(x)\). It also doesn't affect the continuity and differentiability. Moreover, we have \(g(a) = f(a) + k = f(b) + k = g(b)\). Hence, Rolle's Theorem can be applied on \([a, b]\). Thus, the corresponding critical number is c.

Step 2: Determining interval for \(g(x)=f(x-k)\)

For \(g(x) = f(x - k)\), the constant \(k\) shifts the function horizontally but again, it doesn't affect the derivative, hence \(g'(x) = f'(x)\). It also doesn't affect the continuity and differentiability. Moreover, we have \(g(a) = f(a - k)\) and \(g(b) = f(b - k)\). Unless \(a - k = b - k\), \(g(a)\) will not equal \(g(b)\). Rolle's Theorem doesn't immediately apply here. Hence the interval over which Rolle's Theorem can be applied will be difficult to determine without specific values of \(a, b, k\).

Step 3: Determining interval for \(g(x)=f(kx)\)

For \(g(x) = f(kx)\), multiplying the input by \(k\) results in a horizontal scaling of the function. This affects the derivative, as \(g'(x) = k \cdot f'(kx)\). The function is still continuous and differentiable. However, \(g(a) = f(ka)\) and \(g(b) = f(kb)\). Thus, unless \(ka = kb\), \(g(a)\) will not equal \(g(b)\). Once again, Rolle's Theorem doesn't immediately apply here. Hence the interval over which Rolle's Theorem can be applied will be difficult to determine without specific values of \(a, b, k\).