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If g is continuous on a and f(x)=((x^2)-(a^2))g(x). Determine f'(a).

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murciof | Student, College Freshman | (Level 1) eNoter

Posted May 7, 2013 at 3:56 PM via web

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If g is continuous on and f(x)=((x^2)-(a^2))g(x). Determine f'(a).

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oldnick | (Level 1) Valedictorian

Posted May 8, 2013 at 1:46 AM (Answer #2)

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First we not that f(x) is continue too, as product of two functons  `(x^2-a^2)`  and `g(x) ` contiunes.

Now if we suppose that `I ` is the neighbour of `a`  where `g(x)`  is contiunos, since  `(x^2-a^2)` is continue on all over `R`  then in `I`     `f(x)`  is contiunue too. So if exists a limit `b< oo` :

`lim_(x->a) (x^2-a^2)g(x)=b` `=f(a)` 

Derivative is:

`lim_(x->a) (f(x)-f(a))/(x-a)=lim_(x->a) (x^2-a^2)(g(x)-g(a))/(x-a)=`

`=lim_( x->a) (x+a)(x-a)(g(x)-g(a))/(x-a)=``=lim_(x->a) (x+a)(g(x)-b)=` `2a lim_(x->a) (g(x)-b)=0`

Note that  it is not clear that: `lim_(x->a)(x^2-a^2)(g(x)-g(a))/(x-a)=0`

for `g(x)`  is continue on `I` according hypothesis, but not limited.

We have added the hypotesis  `lim_(x->a) (x^2-a^2) g(x)= b`

so that `g(x) ` is dominated by  `(x^2-a^2) ` in `I` .and so make us sure  exists derivative of `f(x)` in `a` by passing the fact we dunno what realy `g(x) ` behaviour is. Actually we know only `f(x)` is contiune in `a` as product of two contiunue functions, nothing short of. 

Indeed let ya see  `g(x)= (log^2(x/a))/(x-a)^2`

then: `lim_(x->a) g(x)=lim_(x->a) (log(x/a) a/x)/(x-a)=`

`=lim_(x->a) a/x xx lim_(x->a) a/x = 1`

We have applied De L'Hospital's rule a twice.

Thus  exists `lim_(x->a) g(x)`  

On the other side:

`lim_(x->a) (x^2-a^2)(g(x)-g(a))/(x-a)^3=lim_(x->a) ((x+a)(log^2(a/x)-1))/(x-a)^2 ` 

does'nt exists  and neither `f'(x)`  

So the  condition of merely limit  isn't enough to assure existence of the limt of product of two funtion , even contiunes,and rtheir relative  derivative existence.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 7, 2013 at 6:18 PM (Answer #1)

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You need to use the definition of derivative at a poin `x = a` , such that:

`f'(a) = lim_(x->a)(f(x) - f(a))/(x - a)`

`f'(a) = lim_(x->a)((x^2 - a^2)*g(x) - (a^2 - a^2)*g(a))/(x - a)`

Reducing duplicate terms yields:

`f'(a) = lim_(x->a)(x^2 - a^2)*(g(x))/(x - a)`

Converting the difference of squares into a product yields:

`f'(a) = lim_(x->a)(x - a)(x + a)*(g(x))/(x - a)`

Reducing duplicate terms yields:

`f'(a) = lim_(x->a)(x + a)*(g(x))`

`f'(a) = 2a*g(a)`

Hence, evaluating the derivative of the function f(x) at x = a, yields `f'(a) = 2a*g(a).`

 

 

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