so that $x$ is $-ve$ and ${I_2},x \ge 0$ so that $x$ is $+ve.$
For ${I_1},f(x) = \int_{ - 1}^x {( - t)dt = - \frac{1}{2}({x^2} - 1)} $.....$(i)$
For ${I_2},f(x) = \int_{ - 1}^x {( - t)dt + \int_{ 0 }^x {( t)dt }}$
$ = - \frac{1}{2}[{t^2}]_{ - 1}^0 + \frac{1}{2}[{t^2}]_0^x = \frac{1}{2}(1 + {x^2})$.....$(ii)$
Hence the function can be defined as the following
$f(x) = \left\{ \begin{array}{l} - \frac{1}{2}({x^2} - 1),{\rm{If}} - 1 \le x < 0\\\frac{1}{2}({x^2} + 1),{\rm{if}}\,\,x \ge 0\end{array} \right.$ ,
$f'(x) = \left\{ \begin{array}{l} - x,\,\,{\rm{if}} - 1 < x < 0\\\,\,0,\,\,{\rm{if}}\,\,x = 0\\\,\,x,\,{\rm{if}}\,\,x > 0\end{array} \right.$
For $f,\,\,L = R = V = \frac{1}{2}$ at $x = 0$, so $f$ is continuous at $x = 0$.
For $f',\,\,L = R = V = 0$ at $x = 0$, so $f'$ is also continuous at $x = 0$.
Thus both $f$ and $f'$ are continuous at $x = 0$ and hence both are continuous for $x > - 1$
$i.e.,$ $x + 1 > 0$..
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