91Ó°ÊÓ

First, let's find the antiderivative for each function: Function \(f_1(x)\): \(f_1(x) = 0, x\leq c\) \(f_1(x) = 1, c<x\) We can find the antiderivative of this function by integrating it over the different intervals. For \(x\leq c\): \(F_1(x) = \int_{a}^{x} f_1(t) dt = \int_{a}^{c} 0 dt = 0\) For \(c<x\): \(F_1(x) = \int_{a}^{x}f_1(t) dt = \int_{a}^{c} 0 dt + \int_{c}^{x} 1 dt = 0 + (x-c) = x-c\) Function \(f_2(x)\) is provided in the exercise as: \(f_2(x) = 0\), if \(x \neq c\) \(f_2(x) = 1\), if \(x = c\) Since \(f_2(x) = 0\) almost everywhere in the interval \([a, b]\), we can integrate over the interval \([a, c)\cup(c, b]\), ignoring the values at point \(c\). For \(x\in[a, c)\cup(c, b]\): \(F_2(x) = \int_{a}^{x} f_2(t) dt = \int_{a}^{x} 0 dt = 0\) Function \(f_3(x)\) is given by: \(f_3(x) = \frac{x-c}{a-c}\), if \(x \leq c\) \(f_3(x) = \frac{x-c}{b-c}\), if \(c<x\) For \(x\leq c\): \(F_3(x) = \int_{a}^{x} f_3(t) dt = \int_{a}^{x} \frac{t-c}{a-c} dt = \frac{(x^2-c^2)/2+c(x-c)}{2(a-c)}\) For \(c<x\): \(F_3(x) = \int_{a}^{x} f_3(t) dt = \int_{a}^{c} \frac{t-c}{a-c} dt + \int_{c}^{x} \frac{t-c}{b-c} dt = \frac{(c^2-c^2)/2+c(c-c)}{2(a-c)} + \frac{(x^2-c^2)/2}{2(b-c)} = \frac{(x^2-c^2)}{2(b-c)}\)

Now, let's analyze the continuity and differentiability of the functions and their antiderivatives at point \(x=c\). Function \(f_1(x)\): As \(f_1(x) = 0\) for \(x\leq c\) and \(f_1(x) = 1\) for \(c<x\), the function is discontinuous at \(x=c\). Antiderivative \(F_1(x)\): For \(x\leq c\), \(F_1(x) = 0\) For \(c<x\), \(F_1(x) = x-c\) As the antiderivative \(F_1(x)\) is continuous at \(x=c\) as it goes from \(0\) to \(0\), it is not differentiable at \(x=c\) because of the jump discontinuity in its derivative \(f_1(x)\) at \(x=c\). Function \(f_2(x)\): As \(f_2(x) = 0\) for \(x\neq c\) and \(f_2(x) = 1\) for \(x=c\), the function is discontinuous at \(x=c\). Antiderivative \(F_2(x)\): For \(x\in[a, c)\cup(c, b]\), \(F_2(x) = 0\). Since \(F_2(x)\) is constant, it is differentiable everywhere in its domain, including at \(x=c\). However, as \(f_2(x) \neq 0\) for \(x=c\), we have that \(f_2(c) \neq F_2'(c)\). Function \(f_3(x)\): The function is continuous at \(x=c\) because both expressions defining it agree that \(f_3(c) = 0\). For \(x\leq c\), \(F_3(x) = \frac{(x^2-c^2)/2+c(x-c)}{2(a-c)}\) For \(c<x\), \(F_3(x) = \frac{(x^2-c^2)}{2(b-c)}\) At \(x=c\), both expressions for \(f_3(x)\) become \(0\), but the derivative of the antiderivative exists and is equal to the function value: \(F_3'(c) = \frac{d}{dx}( \frac{(x^2-c^2)/2+c(x-c)}{2(a-c)} )|_{x=c} = \frac{d}{dx}( \frac{(x^2-c^2)}{2(b-c)} )|_{x=c} = f_3(c) = 0\) Thus, \(F_3(x)\) is differentiable at \(x=c\) and its derivative equals the function value, \(F'_3(c) = f_3(c)\).