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Recent questions tagged continuity-and-differentiability

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GATE CH 2021 | Question: 16
For the function $f(x) = \begin{cases} -x, & x<0 \\ x^2, & x \geq 0 \end{cases}$ ... $x=0$ $f(x)$ is differentiable at $x=0$
For the function $f(x) = \begin{cases} -x, & x<0 \\ x^2, & x \geq 0 \end{cases}$ the $\text{CORRECT}$ statement(s) is/are$f(x)$ is continuous at $x=1$$f(x) $ is different...
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