Proof

[yojoe]

Suppose that f and g are functions which are
(1) differentiable at a=0, and
(2) that f(x)g(x)=x for all real numbers x.

(a) Prove that if a does not = 0 then neither f(a)=0 nor g(a)=0
(b) Use the product rule to show that either f(0) not=0 or g(0) not=0.


(a) Would you prove that with any other number say 1, that neither f(a) nor g(a) exists because there is continuity at 1 rather than at 0?
(b) Same concept apply here too? But then how would you find out what
f(1)=? and g(1)=?

Thanks

 

[pka]

For all x≠0, if a*b=x then neither a nor b can be 0. That shows part a.

Now their product f(x)*g(x)=x for all x. The product has a derivative at x=0.
The derivative of f(x)*g(x)=x must be f’(x)g(x)+f(x)g’(x)=1, WHY?
Is this possible if both f(0) and g(0) equal 0?

 

[Unco]

Suppose that f and g are functions which are
(1) differentiable at a=0, and
(2) that f(x)g(x)=x for all real numbers x.

a) Prove that if a does not = 0 then neither f(a)=0 nor g(a)=0

Would you prove that with any other number say 1, that neither f(a) nor g(a) exists because there is continuity at 1 rather than at 0?

From (2), f(a)g(a) = a

Therefore if a is not zero, then f(a)g(a) is not zero. You can a draw statement from this.