- #1
Felafel
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Homework Statement
Determine for which real values of a,b,c,d this function is differentiable ##\forall x \in \mathbb{R}##:
##f(x):=##
##ax+b ## ## for x\leq1##
##ax^2+c ## ## for 1\leq x \leq2##
##\frac{dx^2 +1}{x} ## ##for x>2.##
The Attempt at a Solution
I know a function is differentiable when the right and left derivatives exist and are equal.So,
##\displaystyle \lim_{x \to 1^-} \frac{(ax+b)-(a+b)}{x-1}=\displaystyle \lim_{x \to 1^+} \frac{(ax^2+c)-(a+c)}{x-1}##
##\displaystyle \lim_{x \to 1^-} \frac{a(x-1)}{x-1}=\displaystyle \lim_{x \to 1^+}\frac{a(x^2-1)}{x-1}##
##a=2a\Rightarrow## a=0
##\displaystyle \lim_{x \to 2^-} \frac{ax^2+c-(4a+c)}{x-2}=\displaystyle \lim_{x \to 2^+} \frac{\frac{dx^2+1}{x}-\frac{4d+1}{2}}{x-2}##
##\displaystyle \lim_{x \to 2^-} \frac{a(x^2-4)}{x-2}=a(x+2)=4a=0##
##\displaystyle \lim_{x \to 2^+} \frac{\frac{dx^2+1}{x}-\frac{4d+1}{2}}{x-2}=0##
##\frac{2dx^2+2-4dx-x}{2x}=0## = ##2dx^2-x(4d+1)+2=0##
##x_{1,2}=\frac{4d+1 \pm \sqrt{16d^2+1-8d}}{4d}##
And so, I'd say ##d=\frac{1}{4}##. Is it correct? Because I have an indeterminate form and I don't think I can use de l'Hospital, as I should know in advance that the limit exists in order to do that. But otherwise the function wouldn't be differentiable.
Also, is it correct to say that b and c can have any finite value, because they "disappear"?
Thank you in advance :)