# how to solve variable in continous function and prove it is differentialable with a function include sinx

Assuming, $$f(x)=\frac{x^2-9}{x^2+2x-3},\;\;\;\text{if}\;\;x<-3$$ and $$f(x)=a\sin(\pi x)+b,\;\;\;\text{if}\;\; x\geq-3$$ $a$ and $b$ are some constants. find $a$ and $b$ if $f(x)$ is continuous everywhere I use left limit =right limit to compute it I find that $$a\sin(-3\pi)+b =3/2$$ but it is the final ans?? how can I solve $a$ and $b$ also, the question also ask me :are any $a$ and $b$ can make $f(x)$ differentiable everywhere? how can I prove it is correct or incorrext, I have no idea about this.

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Your approach -- equating limits -- is correct, but there is something missing in your question. Where did $b$ come from? – glebovg Dec 10 '12 at 8:11
the question said that a and b ar e some constant – user52477 Dec 10 '12 at 8:12
There is not $b$ in the definition of $f$. – glebovg Dec 10 '12 at 8:13
sorry i miss it – user52477 Dec 10 '12 at 8:14
To find $a$ and $b$, evaluate $f$ at some points. – glebovg Dec 10 '12 at 8:15

If $f$ is differentiable at $x=-3$, then the left-sided derivative and right-sided derivative are equal. Then \begin{align} \lim_{h\to +0}\frac{f(-3+h)-f(-3)}{h}&=\lim_{h\to+0}\frac{a\sin(\pi (-3+h))-a\sin(-3\pi)}{h}\\ &=\lim_{h\to +0}\frac{a(\sin(h\pi)\cos(3\pi) -\cos(h\pi)\sin(3\pi)+\sin (3\pi))}{h}\\ &=-a\pi \end{align} and \begin{align} \lim_{h\to -0}\frac{f(-3+h)-f(-3)}{h}&=\lim_{h\to -0}\frac{\frac{h-6}{h-4}-\frac{3}{2}}{h}\\ &=\lim_{h\to-0}\frac{2h-12-3h+12}{2h(h-4)}\\ &=\lim_{h\to-0}\frac{1}{2(4-h)}\\ &=\frac{1}{8}. \end{align} Therefore, we get $a=-\dfrac{1}{8\pi}$ and $b=\dfrac{3}{2}$.