[Solution Library] Let f(x)= \begincases-2 x \text if x<0 , 2 x \text if 0 ≤q x<1 , 1/2(x+1)^2 \text if x ≥q 1endcases Is f(x) is continuous at x=0 ?

Question: Let

$f(x)= \begin{cases}-2 x & \text { if } x<0 \\ 2 x & \text { if } 0 \leq x<1 & \\ \frac{1}{2}(x+1)^{2} & \text { if } x \geq 1\end{cases}$

Is $f(x)$ is continuous at $x=0 ?$
Is $f(x)$ has a derivative at $x=0 ?$ If so, find the value of it.
Calculate $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Calculate $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$
$f\left( 1 \right)$
Is $f(x)$ is continuous at $x=1$ ?
Does $f(x)$ has a derivative at $x=1$ ? If so, find the value of it.

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Deliverable: Word Document

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[Solution Library] Let f(x)= \begincases-2 x \text if x<0 , 2 x \text if 0 ≤q x<1 , 1/2(x+1)^2 \text if x ≥q 1endcases Is f(x) is continuous at x=0 ?

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