If f(x) = begin{cases} int_{0}^{x} (5 + |1-t| | vedclass

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(B) For $x > 2$,$f(x) = \int_{0}^{1} (5 + (1-t)) \, dt + \int_{1}^{x} (5 + (t-1)) \, dt$. Evaluating the integrals: $f(x) = \int_{0}^{1} (6-t) \, dt +

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$f(x)$ is continuous but not differentiable at $x=2$

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(B) For $x > 2$,$f(x) = \int_{0}^{1} (5 + (1-t)) \, dt + \int_{1}^{x} (5 + (t-1)) \, dt$. Evaluating the integrals: $f(x) = \int_{0}^{1} (6-t) \, dt + \int_{1}^{x} (4+t) \, dt = [6t - \frac{t^2}{2}]_{0}^{1} + [4t + \frac{t^2}{2}]_{1}^{x}$. $f(x) = (6 - \frac{1}{2}) + (4x + \frac{x^2}{2} - 4 - \frac{1}{2}) = \frac{11}{2} + 4x + \frac{x^2}{2} - \frac{9}{2} = \frac{x^2}{2} + 4x + 1$. Check continuity at $x=2$: $f(2^-) = 5(2) + 1 = 11$. $f(2^+) = \frac{2^2}{2} + 4(2) + 1 = 2 + 8 + 1 = 11$. Since $f(2^-) = f(2^+) = 11$,$f(x)$ is continuous at $x=2$. Check differentiability at $x=2$: $Lf'(2) = \frac{d}{dx}(5x+1)|_{x=2} = 5$. $Rf'(2) = \frac{d}{dx}(\frac{x^2}{2} + 4x + 1)|_{x=2} = (x+4)|_{x=2} = 2+4 = 6$. Since $Lf'(2) 
eq Rf'(2)$,$f(x)$ is not differentiable at $x=2$.

If $f(x) = \begin{cases} \int_{0}^{x} (5 + |1-t|) \, dt, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}$,then:

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