Vtyjkyuk

Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite Theorem : A function $f:Dto mathbb R$ is continuous in $ain D$ $iff f$ is left and right continuous in $a$. Proof: I firstly thought just to write down the definitions of left and right continuous and then it trivially shows the theorem. But apparently it isn't sufficient. Let $f:Dtomathbb R$ and consider $ain D$. The function $f$ is rightcontinuous in $a iff$ $$(forallepsilongt 0)(existsdeltagt 0)(ale xlt a+deltaRightarrow |f(x)-f(a)|ltepsilon)$$ and left continous $iff$ $a iff$ $$(forallepsilongt 0)(existsdeltagt 0)(a -deltalt xle aRightarrow |f(x)-f(a)|ltepsilon)$$ So I found a proof online on this webpage. My question is there another way to prove this maybe with the use of my definitions? I'd most appreciate it. calculus proof-writing alternative-proof epsilon-delta share | cite | improve this question edited Aug 24 at 11:10 asked Aug 24 at 10:38 Anonymous I 835 1 7 2...

Clash Royale CLAN TAG #URR8PPP up vote 3 down vote favorite For this equation I'm using the following property $$f(x)=e^kx$$ $$f'(x)=ke^kx$$ As well as the product rule $$f(x)=uv$$ $$f'(x)=u'v+uv'$$ I factorize $x$ on $e$'s exponent and then use the first property to differentiate: $$e^7x^3-frac53=e^x(7x^2)-frac53=7x^2*e^7x^3-frac53$$ Is his fully differentiated? Or do I have to apply the product rule to $7x^2$? Any other steps I'm missing? derivatives share | cite | improve this question edited Aug 24 at 10:43 asked Aug 24 at 10:38 Pablo 333 12 $7x^2$ isn't a constant. – poyea Aug 24 at 10:42 1 Ahem. $uv$ is a product, not a quotient. – Sean Roberson Aug 24 at 10:42 @SeanRoberson lmao my bad – Pablo Aug 24 at 10:43 add a comment  |  up vote 3 down vote favorite For this ...