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

#1**+1 **

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

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

.Rom Oct 17, 2018

#2**0 **

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

#3**0 **

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

#4**0 **

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.)}

GA

Guest Oct 20, 2018