“Switching” of a continuous function

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Let $f, g: [0,1] rightarrow mathbbR$ be continuous functions and consider the continuous function $h(x) = maxf(x), g(x)$. Suppose that at $0$, $f$ is "active" and at $1$, $g$ is active. That is $h(0) = f(0) neq g(0)$, and similarly at $x = 1$. Is it true that there must exist some point $t in [0,1]$ such that $h$ "switches" from $f$ to $g$? If not I think this should violate continuity but I don't see exactly how.




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    Let $f, g: [0,1] rightarrow mathbbR$ be continuous functions and consider the continuous function $h(x) = maxf(x), g(x)$. Suppose that at $0$, $f$ is "active" and at $1$, $g$ is active. That is $h(0) = f(0) neq g(0)$, and similarly at $x = 1$. Is it true that there must exist some point $t in [0,1]$ such that $h$ "switches" from $f$ to $g$? If not I think this should violate continuity but I don't see exactly how.




    continuity




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      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $f, g: [0,1] rightarrow mathbbR$ be continuous functions and consider the continuous function $h(x) = maxf(x), g(x)$. Suppose that at $0$, $f$ is "active" and at $1$, $g$ is active. That is $h(0) = f(0) neq g(0)$, and similarly at $x = 1$. Is it true that there must exist some point $t in [0,1]$ such that $h$ "switches" from $f$ to $g$? If not I think this should violate continuity but I don't see exactly how.




      continuity




      share|cite|improve this question











      Let $f, g: [0,1] rightarrow mathbbR$ be continuous functions and consider the continuous function $h(x) = maxf(x), g(x)$. Suppose that at $0$, $f$ is "active" and at $1$, $g$ is active. That is $h(0) = f(0) neq g(0)$, and similarly at $x = 1$. Is it true that there must exist some point $t in [0,1]$ such that $h$ "switches" from $f$ to $g$? If not I think this should violate continuity but I don't see exactly how.





      continuity





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      asked Aug 6 at 19:59









      user395788

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          Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...






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          • Damn, I figured there was probably a standard trick to show this easily. Thanks.
            – user395788
            Aug 6 at 20:03










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          Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...






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          • Damn, I figured there was probably a standard trick to show this easily. Thanks.
            – user395788
            Aug 6 at 20:03














          up vote
          1
          down vote



          accepted










          Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...






          share|cite|improve this answer





















          • Damn, I figured there was probably a standard trick to show this easily. Thanks.
            – user395788
            Aug 6 at 20:03












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...






          share|cite|improve this answer













          Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...







          share|cite|improve this answer













          share|cite|improve this answer



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          answered Aug 6 at 20:01









          Nicolas FRANCOIS

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          • Damn, I figured there was probably a standard trick to show this easily. Thanks.
            – user395788
            Aug 6 at 20:03
















          • Damn, I figured there was probably a standard trick to show this easily. Thanks.
            – user395788
            Aug 6 at 20:03















          Damn, I figured there was probably a standard trick to show this easily. Thanks.
          – user395788
          Aug 6 at 20:03




          Damn, I figured there was probably a standard trick to show this easily. Thanks.
          – user395788
          Aug 6 at 20:03












           

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