Let f be a function defined below. Discuss the continuity of f at x=−3,x=0,

Jaya Legge

Jaya Legge

Answered question

2021-10-27

Let f be a function defined below.
Discuss the continuity of f at x=−3,x=0, and x=1. If discontinuous, give the type of discontinuity.
f(x)={|x2+3xx+3|x0,x3x+1if0<x<1xifx1

Answer & Explanation

liannemdh

liannemdh

Skilled2021-10-28Added 106 answers

Given
The function as
f(x)={|x2+3xx+3|x0,x3x+1if0<x<1xifx1
For limx0f(x)
f(x)=x+1 if =0 Since, it is clear that function has changed its definition three times here i.e. at
x=0
x=1 and
x=3
So, let us first discuss the continuity at x=0
Check for if, limx0f(x)=f(0)
Here, f(x)=|x2+3xx+3| if x0, x3
f(0)=02+3×00+3 if x0, x3
f(0)=02+3×00+3 if x0,x3
f(0)=0
For limx0+f(x)
f(x)=x+1 if 0<x<1
Therefore,
limx0+f(x)=limx0+x+1=0+1=1
and,
For limx0f(x)
f(x)=|x2+3xx+3|
lim0f(x)=limx0|x2+3xx+3|
=(x23xx+3)
=(02+3×00+3)
=03
=0
Therefore, limx0f(x) does not exist.
Hence, f(x) is not continuous at x= 0 and since left hand side and right hand side limit exists separately but not equal , so its a Jump type of discontinuity.
Now for, x=1
limx

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