1is this fmultilpied y g or fog ?? sry for silly doubt
If f (x) is continuous and g (x) is discontinuous at x = a are then f (x) +/-- g (x) is discontinuous at x = a
now is this always true or depends on the case???
i'll give an eg. f(x) = 1/x ; g(x) = 1/x ; h(x) = f(x) - g(x);
then is h(x) cont. or discont at x=0;
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17 Answers
62in that function is not defined at x=0
I can given another example where it is discontinuous.
f(x)=[x]
g(x)=1/[x]
both the above are defined for x>3
f(x).g(x) is 1 always
which is continuous at x=integer.
21hmmmmm....... k thnx.......
and sir wat abt the eg provided abov??
is h discont at x=0
62F(x).G (x) is not necessarily discontinuous.
trivial example F(x)=0 for all x
and G(x) is any discontinuous function.
21oh k fine so f + g ka to confirm ho gaya.....
sir can u pl. confirm it 4 f(X)*G(X) = H(X)
ie if F(X) = X ; G(X) = 1/X
11It is discontinuous.
Other similar examples are :
y=sinx.cotx at x=0
y= cosx.tanx at x=Ï€/2
tanxcotx at x=0 and π/2
21hey ani u may call me by "any" name dude naam mein kya rakha hai?? kam dekho!!! LOL...
hey guyz dont u think this is strange coz we are almost defining h(x) = 1 but in a diff meth..... then y shud it not b cont?????? i think its a matter of CONVENTION coz in case of h(x) = x/x ; limit exists......
BUT YEAH maybe i too feel it wud be discont.....
11Don't care whether it is f+g or f*g........If the elements are not defined then function is not defined.
Tapan (if i may call u that): second h(x) is also not defined
21but yes eureka in da "f+g" post of urs i feel u r spot on!!!
then how f*g???
21its a really controversial questn that i hav posted as both sides seem correct and its da questn wat it is taken as in PURE MATHS.....
another similar thing happens wen u take f(x) = x ; g(x) = 1/x; h(x) = f(x) * g (x);
now will this be cont at x=0????????// [7]
11There is a test in this website just involving this concept. All elements of the function must be defined.....even if they cancel each other out
24if f is continous and g is discontinuous then f+g is never continuous.....[1]
11YA SUBASH.......it is 0 for all x≠0
For x=0, both f(x) and g(x) are induvidually not defined.......so h(x) is not defined [4]
11hows that ani func is zero always
maybe it holds for non zero functions