If f(x) = int_{-1}^x |t| , dt,x ge -1,then | vedclass

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(A) Let us divide the interval into two sub-intervals: $I_1: -1 \le x < 0$ and $I_2: x \ge 0$.For $I_1$,$f(x) = \int_{-1}^x (-t) \, dt = -\frac{1}{2}[

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$f$ and $f'$ are continuous for $x + 1 > 0$

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(A) Let us divide the interval into two sub-intervals: $I_1: -1 \le x For $I_1$,$f(x) = \int_{-1}^x (-t) \, dt = -\frac{1}{2}[t^2]_{-1}^x = -\frac{1}{2}(x^2 - 1) = \frac{1}{2}(1 - x^2)$.For $I_2$,$f(x) = \int_{-1}^0 (-t) \, dt + \int_0^x (t) \, dt = -\frac{1}{2}[t^2]_{-1}^0 + \frac{1}{2}[t^2]_0^x = -\frac{1}{2}(0 - 1) + \frac{1}{2}(x^2 - 0) = \frac{1}{2}(1 + x^2)$.Thus,$f(x) = \begin{cases} \frac{1}{2}(1 - x^2), & -1 \le x Taking the derivative,$f'(x) = \begin{cases} -x, & -1  0 \end{cases}$.At $x = 0$,$\lim_{x 	o 0^-} f(x) = \frac{1}{2}(1 - 0) = \frac{1}{2}$ and $\lim_{x 	o 0^+} f(x) = \frac{1}{2}(1 + 0) = \frac{1}{2}$. Since $f(0) = \frac{1}{2}$,$f$ is continuous at $x = 0$.For $f'(x)$,$\lim_{x 	o 0^-} f'(x) = 0$ and $\lim_{x 	o 0^+} f'(x) = 0$. Since $f'(0) = 0$,$f'$ is continuous at $x = 0$.Therefore,both $f$ and $f'$ are continuous for all $x > -1$,i.e.,$x + 1 > 0$.

If $f(x) = \int_{-1}^x |t| \, dt$,$x \ge -1$,then

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