NompsypeFeplk

2021-11-15

Discuss the continuity of the function

on the open interval $0 and on the closed interval $0\le x\le 2$

Volosoyx

Consider the function,

To check function f(x) is continuous on open interval 0 In open interval 0 The function $f\left(x\right)={x}^{2}-3x$ is polynomials of degree 2 and polynomials function are continuous function.
Therefore f(x) is continuous on interval (0,2)
To check function f(x) is continuous on closed interval $0\le x\le 2$
If function f(x) is continuous at points x=2 then,
$\underset{x\to {2}^{-}}{lim}f\left(x\right)=\underset{x\to {2}^{+}}{lim}f\left(x\right)=f\left(2\right)$
To calculate the left hand limit,
$\underset{x\to {2}^{-}}{lim}f\left(x\right)=\underset{x\to {2}^{-}}{lim}\left({x}^{2}-3x\right)$
$={\left(2\right)}^{2}-3\left(2\right)$
$=4-6$
$=-2$
To calculate the right hand limit,
$\underset{x\to {2}^{+}}{lim}f\left(x\right)=\underset{x\to {2}^{-}}{lim}\left(2x+4\right)$
$=2\left(2\right)+4$
$=8$
$\underset{x\to {2}^{-}}{lim}f\left(x\right)\ne \underset{x\to {2}^{+}}{lim}f\left(x\right)$
$-2\ne 8$
Here left hand limit is not equal to right hand limit.
Hence, the function is discontinuous.

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