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Let f(x) be a derivable function defined over the set of real numbers such that

$$f(f(x))=k(x^7+2x),$$ $k$ is not equal to zero. Then we have to prove $f(x)$ is either increasing or decreasing .

I am not getting any try. Can anybody help me here.

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    Differentiate f, what do you obtain? – 2017-02-19
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    Is $f(f(x))=k(x^7+2x)$ or $f(x)=k(x^7+2x)$? – 2017-02-19
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    try to take derivative of this equation using the chain rule. What can you say about points where the derivative is zero? – 2017-02-19
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    @stud_iisc the first one is correct – 2017-02-19
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    @RyArazi I could not understandf how – 2017-02-19
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    the chain rule gives you: $f'(x) \cdot f'(f(x)) = k(7x^6+2)$. what happens when f'(x) =0? is there a solution? – 2017-02-19
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    @RyArazi no there is no solution – 2017-02-19
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    And what happens if for some $x_1,x_2$ you will find $f(x_1)>0, f(x_2)<0$? What Darboux's theorem says in this case? – 2017-02-19
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    @RyArazi when that happens it means k=0 or $f'(f(x))=infinity$ – 2017-02-20
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    @laflaca after that – 2017-02-20
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    you wrote that $k\ne0$ in your question. Also $f$ is derivable and that means $f'(y)\in \mathbb{R}$ for every real $y$ (and for $f(x)$). So what does it means? – 2017-02-20
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    @RyArazi that means it can be increasing or decreasing . But it cannot be constant . Am I correct – 2017-02-20
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    it means more than that - it means that $f'$ can't change it's sign because in each change it will must go through a point in which $f'(x)=0$! – 2017-02-20

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