Question

can someone please explain this problem for me in calc?

Answer 1

@ganeshie8

Answer 2

u know how to find continuity of a function

Answer 3

not skilled

Answer 4

cn u draw its graph

Answer 5

i need to draw a graph for this problem?..

Answer 6

not necessary

Answer 7

u find limit at x=1

Answer 8

u know how to find limit

Answer 9

bt its not a pure quadratic function

Answer 10

mondona u know how to find limits

Answer 11

yes

Answer 12

then find left hand and right hand limt at 1

Answer 13

for left hand limit use eq 3-x

Answer 14

ok nevermind im confused, ill ask my teacher

Answer 15

ok as u wish

Answer 16

:(

Answer 17

if u got the answer so u may close this Q! THANKS

Answer 18

you have a couple of equations to set up

Answer 19

you need to set the left limit equal to the right limit as x approaches 1
you also need to set the left limit equal to the right limit as x approaches 2
this should give a system of equations to solve

Answer 20

that will give you a and b such that it makes the function continuous

Answer 21

or since they already give you a and b
you can still do those things up there to check for continuity

Answer 22

as x approaches 1, 3 -x approaches 2
as x approaches 1, 2x^2 + 3x approaches 5

so there is a discontinuity here ( I think!)
- such a long time since i did these .....

Answer 23

my memory of this is a bit vague i'm afraid

Answer 24

looks good @cwrw238
If the left limit doesn't equal the right limit as x approaches 1,then f is discontinuous at x=1.
You must have
f(1) exists
limx->1 f(x) exists
and finally f(1)=limx->1 f(x)
the 2nd one exists if left limit equals right limit as x approaches 1