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I want to prove that $x< y \implies f(x) < f(y) \forall x,y \in [0,\infty)$ for the function $f(x)=\exp(x)-\sin(\exp(-x))$ without using derivatives.

I know that $e^x$ is increasing but $\sin(e^{-x})$ is decreasing and $\sin(e^{-y})< \sin(e^{-x}) \leq \sin(1) <1$, so I think this doesn't help me if I want to show that $f(x)-f(y)>0$ for $x>y$.

So I am totally stuck there (and by similar tasks) and I am not sure how to tackle such problems. How can I continue this proof?

Quaeram
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1 Answers1

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As I suggested in my comment, we can rewrite our function in the form: $$f(x)=e^x-\sin(e^{-x})=e^x+\underbrace{(-\sin(e^{-x}))}_{g(x)}$$ $g(x)$ is a composite function. $e^{-x}$ is a decreasing function in $(0,1]$. $\sin(t)$ is increasing in $I^+(0)$, so $g(x)$ is increasing in $[0,+\infty)$, as we can see from $g'(x)$: $$g'(x)=-\cos(e^{-x})\cdot (-1)\cdot e^{-x}=\cos(e^{-x})\cdot e^{-x}>0$$

$f(x)$ is a sum of two increasing function, so it's itself increasing.

Matteo
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  • @Quaeram: if this answer has been useful, let me know smahing that +1 and accepted answer button. Thanks. – Matteo Feb 09 '22 at 14:29
  • It was very useful for me and I would have clicked +1 if I had enough reputation, so I will do it later ;) – Quaeram Feb 09 '22 at 15:06