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Let $f\colon[1,8]\to \Bbb{R}$ be a continuous function such that $f(1)=4$ and $f(8)=3$.

Show that exists $x_0 \in [1,8]$ such that $$f(x_0)= \sqrt[3]{x_0}+x_0.$$

How can I start this?

DMcMor
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Anıl Geyik
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    Set $g(x):= f(x) - x^{1/3} - x$. Consider what happens if you substitute in $x=1$ and $x=8$. – Soby Apr 15 '21 at 19:05

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Consider the function $g(x) = f(x) - \sqrt[3]x - x$ and notice that $g$ is also continuous in $[1,8]$.

Further notice that $g(1) = 4 -1 -1 = 2$ and $g(8) = 3 - 2 - 8 = -7 $. In particular, $ \ g(1) > 0 > g(8)$.

Then, by the Intermediate value theorem, there is $x_0 \in ]0,8[$ such that $g(x_0) = 0$.

Babado
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  • Thank you so much I get it – Anıl Geyik Apr 15 '21 at 19:08
  • You're welcome. In these kind of problems where you have to find a value which satisfies some condition, you usually define an auxiliary function and apply the intermediate value theorem. – Babado Apr 15 '21 at 19:10