Discuss the continuity of the functionf(x)=\begin{cases}x^2-3x&if\ x<2\\

NompsypeFeplk

NompsypeFeplk

Answered question

2021-11-15

Discuss the continuity of the function
f(x)={x23xif x<22x+4if x2
on the open interval 0<x<2 and on the closed interval 0x2

Answer & Explanation

Volosoyx

Volosoyx

Beginner2021-11-16Added 10 answers

Consider the function,
f(x)={x23xif x<22x+4if x2
To check function f(x) is continuous on open interval 0 In open interval 0 The function f(x)=x23x 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 0x2
If function f(x) is continuous at points x=2 then,
limx2f(x)=limx2+f(x)=f(2)
To calculate the left hand limit,
limx2f(x)=limx2(x23x)
=(2)23(2)
=46
=2
To calculate the right hand limit,
limx2+f(x)=limx2(2x+4)
=2(2)+4
=8
limx2f(x)limx2+f(x)
28
Here left hand limit is not equal to right hand limit.
Hence, the function is discontinuous.

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