I've got an exercise to do and I don't really know what to do.
Exercise : We've got function $f$, where $f(a) = 0$ and $f'(a)$ exists. Also we got function $g$ which is continuous. Does exist $(f-g)'(a)$? Explain it.
My opinion is that exists, but I've got no idea how should I explain it. Some help? Thank you.
$\endgroup$ 13 Answers
$\begingroup$$f(a)=0$ and $\exists f'(a)$.
The expression : $(f-g)'(a)$ is : $f'(a) - g'(a)$.
You cannot say that this derivative exists, if $g$ is not differentiable at $a$. In your question's body, you have only stated that $f$ is differentiable and that $g$ is only continuous. Thus, no, you cannot say generally that $\exists (f-g)'(a)$.
$\endgroup$ 5 $\begingroup$Not true, example: take $f=0$, $g=|x|$ and $a=0$.
$\endgroup$ 1 $\begingroup$Is $g$ differentiable? That information is not given. It is possible for $g$ to be continous but not differentiable.
Anyway differentiation is "linear", in that $(f-g)'(a)$ = $f'(a) - g'(a).$ So, if $g'(a)$ exists (and $f'(a)$ exists which is given) then $(f-g)'(a)$ exists.
But if $g'(a)$ does not exist, then $(f-g)'(a)$ does not exist.
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