since through the first hints, f(x) is continuous at x=0, I know that b=0, and from the second hint that derivative of f(0) is also equal to 0 so is that a and b equal to 0, but it seems that not correct.

Guest Oct 17, 2018

\(f(x) \text{ will be continuous if the each of the individual parts is continuous }\\ \text{and the value of those parts agree at }0 \\ \text{each part is composed of continuous functions so we just have to check values at 0}\\ f(0^-)= a \tan^{-1}(0) + b = b \\ f(0^+) = \lim \limits_{x \to 0}~\dfrac{\ln(1+\sin(x))}{x}\\ \text{Using L'Hopital's rule we take the derivatives of numerator and denominator and look at the ratio} \\ \left .\dfrac{d}{dx} \ln(1+\sin(x)) = \dfrac{\cos(x)}{1+\sin(x)} \right|_{x=0} = 1\\ \dfrac{d}{dx}=1 \text{ and thus the limit is }\\ f(0^+) = \lim \limits_{x \to 0}~\dfrac{\ln(1+\sin(x))}{x} = \dfrac 1 1 = 1 \\ \text{and thus }b=1\)



\(\text{The value of the derivative at 0 must also agree from both sides} \\ \text{I'm going to let you work this bit out}\\ \text{Take the derivatives of both terms of f(x) }\\ \text{one can be immediately evaluated at 0, the other you'll have to take the limit as }x\to 0 \\ \text{set the value }a \text{ such that the two values agree}\)

Rom  Oct 17, 2018

hello , bro,

May  know why you do derivative for ln(1+sinx)/x but ignore the derivative for atan^-1x+b???

Guest Oct 20, 2018

Rom may never see your question.

If you used his name in the first line you would have more hope of him seeing it.

OR  If you became a member you could prod him with a polite private message.

Melody  Oct 20, 2018

Rom explains why in his concluding paragraphs.


“I’m going to let you work this bit out …”


Rom peeled and predigested the first banana for you. You are a big monkey now, so you need to learn how to peel and eat bananas on your own volition. Otherwise, you will never grow up to be big and smart. (Not that there is much hope for the smart part.)



Guest Oct 20, 2018

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