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Given $f$ and $|f|$ are continuous on some interval, show that $e^{-\sqrt{|f|}}$ is also continuous.

I am able to show that $e^{\sqrt{|f|}}$ is continuous. However, with the minus sign on exponential, i don't see why this new function is continues given $\epsilon$ small enough, it just blows up.

79999
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  • If you can show it without the minus sign, it shouldn't be much harder. What is your working out? Also, do you know the result that the composition of continuous functions is also continuous? This would be useful in this case. – JKL May 02 '22 at 01:23
  • the one without minus is trivial, isn't it? oh, i missed the $|f|$ is also continuous in the given condition. the square root function is continuous, then raising the exponential also satisfies the epsilon-delta criteria. However, with the minus, it's like doing $\frac{1}{e^{\epsilon}}$ – 79999 May 02 '22 at 01:38
  • I'm still a bit confused about what your approach is. Maybe if you write down how your "trivial" argument proceeds (I am assuming you are doing an $\epsilon$-$\delta$ argument), it might be clearer what is going on. – JKL May 02 '22 at 02:04
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    basically follows this: https://math.stackexchange.com/questions/605468/if-f-is-continuous-then-is-sqrt-f-continuous – 79999 May 02 '22 at 02:30
  • Right, @79999 , you’re just composing two continuous functions. – Lubin May 02 '22 at 02:43
  • If you define: $$A(x) = |x|\S(x) = \sqrt x, x \ge 0\N(x) = -x\E(x) = e^x$$ then your function is $E\circ N \circ S \circ A \circ f$. Since each function is continuous everywhere on its domain, and each is defined on the entire image of its predecessor, and since the composition of continuous functions is continuous, your function is continuous everywhere. That $|f|$ is continuous follows immediately from the continuity of $f$ and of $A$. A separate assumption for it was not necessary. – Paul Sinclair May 02 '22 at 16:11

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