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Let $Y=f (x)$ be an implicit equation satisfying the relation $f (|2^x - 1|)-2f (|2^{-x} -1|)=3x-2$. where $x \in \mathbb{R}.$

If $f_1 (x)$ and $f_2 (x)$ are explicit functions obtained from the equations whose domains are $[0,1)$ and $[0,\infty)$ respectively. Then

A) Range of $f_1 (x)$ and $f_2 (x)$ are ?

B) Discuss the derivability of $f_1 (x)$ and $f_2 (x).$

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    Sorry of inconvenience in the question it is Y- f (x) =0 is the implict function – 2017-01-27
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    Dear steve, on this site we use MathJaX to format our maths. [Here](http://meta.math.stackexchange.com/q/5020/145141) you can find a basic tutorial. – 2017-01-27
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    I am new to this .. so I don't know how to use mathjax. I hope the question can be understood . – 2017-01-27
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    I have improved your text placing in particular all formulas between dollar signs (think to it next time). Check the changes. Besides what have **you** done on the subject ? – 2017-01-27
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    Thank you sir.. I will take care of it next time . – 2017-01-27
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    Can any try this ... I m not getting it I just simplified implict equation by considering -x . – 2017-01-27
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    No body gave answer.... ? – 2017-01-27
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    Eric Wofsey sir ..... can u please try it – 2017-01-27
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    @JeanMarie can you please try it . – 2017-01-28

1 Answers 1

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Partial solution.

Changing $x$ into $-x$ (you say you have had this idea) gives:

$$\tag{1} f (|2^{-x} - 1|)-2f (|2^{x} -1|)=-3x-2.$$

Let $u:=f (|2^{x} - 1|)$ and $v:=f (|2^{-x} - 1|)$.

The initial equation can be written: $u-2v=3x-2$,

The second equation, (1), can be written: $v-2u=-3x-2$.

Solving this system, gives:

$$\tag{2}5u=9x+4 \ \iff \ \ u=f(|2^x-1|)=\dfrac{9x+4}{5}$$

Set $X:=2^x-1$. This is equivalent to $X+1=2^x \ \iff \ x=\ln(X+1).$

Plugging these expressions (of $x$ and $X$) into (2) gives:

$$f(|X|)=\dfrac{9\ln(X+1)+4}{5}$$

From this expression, I'm confident that you can proceed.

Remark: I think, without being certain (the text is not clear), that $f_1(X):=f(X)$ and $f_2(X):=f(-X).$