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Let f(x) = (ax+sin2x)/3x for x>0

       a(x+1) for x=0

       ae^bx for x<0

Questions:

a) Determine all the values of a and b for which f is a continuous function

b)Determine whether there are values of a and b for which f is a differentiable function.

So using the properties of continuity by first equation the limit of x->0+ to x->0- based on the fact that the function is continuous , I found that the value of a is 1 then since the function is continuous , I can also conclude limit of x->0- will be equal to f(0) and hence computer the value of b as 0.

Hence based on that, I concluded that the for the corresponding values the given function is continuous. I know that continuity does not imply differentiability. How do I go about computing the answer for the second question? Any form of help or hints are appreciated.

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    f (0)=0 so lim x+ f (x)= (a/3)+sin2x/3x = a so a = (3/2)lim sin 2x/3x. b can be anything to be continuous. – 2017-02-24
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    according to my computation, f(0) = 1. – 2017-02-24
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    Set $b=100$. Then $\lim_{x\rightarrow0^{-}}\exp{(100x)}=\exp{(0)}=1$. $b$ can be anything for continuity. – 2017-02-24
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    I meant to type f(0) = a. That was a typo. so lim (ax + sin 2x)/3x = a/3 + sin 2x/3x = a. So a = 3/2 \lim sin 2x/3x. which by lhopital is 1. lim of ae^bx is a so b could be anything as for as continuity is concerned. To be continuous lim- f'(x) = be^bx = lim+ f'(x) = ]2cos 2x*3x - 3(a + sin 2x)]/9x^2. – 2017-02-24

1 Answers 1

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You need to calculate right derivative of your function, namely, of $\frac{ax+\sin(2x)}{3x}$ at $0$, and set it equal to the left derivative at $0$. You already know $a=1$. You need to calculate $\lim_{x\rightarrow0^{+}}\frac{\partial}{\partial x}\frac{x+\sin(2x)}{3x}$, for which I am getting $0$. The left derivative at $x=0$ equals $b\exp(b\cdot0)=b$.