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Question

Show that if $f\colon\mathbb R\longrightarrow\mathbb R$ is a derivable function then $g(x)=f(x)+f'(x)$ has the property of intermediate values.

And if we consider $h(x)=f(x)+f'(x)+f''(x)$ ?

Attempt

Since $f$ is differentiable then it is continuous, if $f'$ is continuous then $f+f'$ is continuous and by the intermediate value theorem $g$ has the property. But what if $f'$ is not continuous ? Does Darboux's theorem guarantee that $f'$ has the property ?

Pierre
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1 Answers1

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Since $f$ is continuous, it has a primitive $F$. $F + f$ is derivable and its derivative is $f + f'$. Since it is a derivative, it has the IVT property.

Same for $f + f' + f''$ with $F + f + f'$.

Jean-Armand Moroni
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  • Can you get a resolution without primitives? – Pierre Jul 11 '22 at 10:21
  • @PedroCosta That's a good question. We could try proving that a Darboux function + a continuous function is a Darboux function. Which would be more general than using a primitive, since some Darboux functions are not a derivative. I'll give it a try. – Jean-Armand Moroni Jul 11 '22 at 16:30
  • Thanks. I'll give a try too. Good job. – Pierre Jul 11 '22 at 16:35
  • @PedroCosta Actually it is false; cf. https://math.stackexchange.com/questions/2659170/is-the-sum-of-a-darboux-function-and-a-continuous-function-darboux – Jean-Armand Moroni Jul 11 '22 at 16:38
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    I think in let $h(x)=e^xg(x)$ in $I\subset \mathbb R$. $h(x)=j'(x)$, where $j(x)=e^xf(x)$. $h$ have a property. As $g(x)=e^{-x} h(x)$, $g$ have a IVT. But $g(x)=f(x)+f'(x)$ – Pierre Jul 11 '22 at 16:42
  • But in this answer exist a little problem. Is who proof that if $f$ has the IVT then so does $f^{-1}$. That's true? – Pierre Jul 11 '22 at 16:47
  • $f^{-1}$ has the IVT property is equivalent to: $f$ domain is an interval. But how do you prove that $g$ has IVT? The product of a Darboux function by a continuous function is not necessarily Darboux. – Jean-Armand Moroni Jul 11 '22 at 17:12
  • You are absolutely right, I had also noticed that detail (great detail). It seems to me impossible not to even use the notion of primitive or something similar. – Pierre Jul 11 '22 at 17:42
  • By the way, the fact that $f^{-1}$ has the property IVT proves that if $f$ is not defined in an interval then can we guarantee that $f^{-1}$ is not continuous ? – Pierre Jul 11 '22 at 17:56
  • Yes, if $f$ is defined on a domain that is not an interval, then $f^{-1} is not continuous. – Jean-Armand Moroni Jul 11 '22 at 23:47
  • Right, Thanks you. – Pierre Jul 12 '22 at 07:18