anonymous
  • anonymous
Let g(x)=x f(x) for all x belonging to R Show that if f is continuous at x=0 then g is derivable at x=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
derivable or differentiable? careful with the jargon :P
anonymous
  • anonymous
derivable ! copied from my buk ! :p
experimentX
  • experimentX
i think both means the same

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anonymous
  • anonymous
Thats ok ! Solve d prob ! :p
anonymous
  • anonymous
it definitely does not mean the same. if that's a calculus book it's a mistake.
anonymous
  • anonymous
malcom---its derivable ! No prob wd d jargon -solve it plzzz !
perl
  • perl
Woh Hansane wali baat thi anjali
perl
  • perl
use definition of continuity
anonymous
  • anonymous
I dint gt u ? Ur nt indian rite ? And plz no hints fr ques - straightforward explanatory ans ! I would be really obliged ! :p
perl
  • perl
anjali, mai apne khaane kaa anand le rahaa hoon.
perl
  • perl
anjali, one minute please
experimentX
  • experimentX
somewhere i had read .. the product of two continuous function is a continuous function is a continuous function ... somehow i can't find it
anonymous
  • anonymous
Hmmm, ok , search , then let me know.
anonymous
  • anonymous
And perl , whose helping you with hindi ?
anonymous
  • anonymous
:P :D
perl
  • perl
anjali, hello
anonymous
  • anonymous
heloo
perl
  • perl
mujhe pratham puraskaar milne kii aashaa hai , medal?
anonymous
  • anonymous
ohk-now u r indian - and dont lie abt dt ! gt ya !
anonymous
  • anonymous
:p
perl
  • perl
ok
anonymous
  • anonymous
wt ok ???
perl
  • perl
no problem
perl
  • perl
Mera naam vivckenanda hai
perl
  • perl
anjali, if f(x) is continuous at x = 0, then lim f(x) = f(0) as x ->0
anonymous
  • anonymous
anyways, plzzzz ans my maths query , i gtg , cya ltrz !
perl
  • perl
the answer , he is right
perl
  • perl
First let's write over g(x) = x * f(x) as g(x) = h(x) * f(x), where h(x) = x We know that h(x) = x is continuous at x = 0. And we know that f(x) is continuous at x = 0 because that is given. We know that the product of continuous functions at a point is continuous at the point (theorem on continuity). So g(x) is continuous at the point. how do we deduce that g(x) is differentiable?
anonymous
  • anonymous
Limit[ (g(x)-g(0))/(x-0), x->0]= Limit[ (xf(x)-0)/(x-0), x->0]=Limit[ (xf(x) )/x, x->0]=Limit[ f(x), x->0]=f(0) From the definition, g is differentiable at - and g'(0)=f(0)
anonymous
  • anonymous
From the definition, g is differentiable at 0 and g'(0)=f(0)
perl
  • perl
what definition of derivative you used >
perl
  • perl
you used lim f(x) - f(a) / ( x - a )
experimentX
  • experimentX
seems alright to me
anonymous
  • anonymous
If Limit[ (g(x) -g(b))/(x-b), x->b]=L exists and finite then g is differentiable at x =b and g'(b)=L
perl
  • perl
Ok let me spruce it up a bit. g is differntiable at x = 0 if ... ok right lim g(x) - g(a) ) / ( x - a) exists ,
perl
  • perl
if the limit exists then it must necessarily be finite
perl
  • perl
ok so you demonstrated that lim [ g(x) - g(0) ] / ( x - 0 ) exists as x-> 0 . the limit of that is the same as the limit of f(x) as x->0 , and we know that exists since f is continuous
anonymous
  • anonymous
Some people may consider Infinity as a limit too. So we have to avoid this possibility.
anonymous
  • anonymous
so you demonstrated that lim [ g(x) - g(0) ] / ( x - 0 ) exists as x-> 0 . the limit of that is the same as the limit of f(x) as x->0 , and we know that exists since f is continuous The answer is yes.
perl
  • perl
oh , good point
perl
  • perl
about infinity
perl
  • perl
but we didnt actually ask for what g'(0) is, so that was extra information
perl
  • perl
actually you can do this proof with a weaker condition, f(x) limit exists at x = 0

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