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### Answers (1)

Answer:

everywhere continuous.

Hint:

Constant and identity function is everywhere continuous.

Given:

$f(x)=\left\{\begin{array}{rr} -2 & x \leq-1 \\ 2 x & -1

Explanation:

Constant and identity function are everywhere continuous. So, we only have to check at x = -1, x = 1 (end points) -(1)

At  x = -1

\begin{aligned} &\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1}-2=-2 \\ &f(-1)=-2 \\ &\lim _{\left(x \rightarrow 1^{+}\right)} f(x)=\lim _{x \rightarrow 1^{-}} 2 \times x \\ &=2 \times-1=-2 \end{aligned}

As

$\begin{gathered} \lim _{x \rightarrow-1} f(x)=f(-1) \\ f(x)_{\text {is continuous at }} x=-1-(2) \end{gathered}$

At x = 1

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 2 x=2 \times 1=2 \\ &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 2=2 \\ &f(1)=2 \end{aligned}

As

$\lim _{x \rightarrow 1} f(x)=f(1)-(3)$

So from (1), (2) & (3)

f(x) is everywhere continuous.

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