Let's consider the continuous function f : R × [ a , b ] → R Such that f ( x , a ) &gt; 0 for all x and there exists x b with f ( x b , b ) ≤ 0.Then there exists c ∈ [ a , b ] such that f ( x , c ) ≥ 0 for all x and a real value x c c with f ( x c , c ) = 0.Intuitively this seems true, but I woludn't know how to prove it.

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Let&#039;s consider the continuous function  f  :      R    ×  [  a  ,  b  ]  →      R  Such that   f  (  x  ,  a  )  &amp;gt;  0 for all   x and there exists       x    b   with   f  (      x    b    ,  b  )  ≤  0.Then there exists   c  ∈  [  a  ,  b  ] such that   f  (  x  ,  c  )  ≥  0 for all   x and a real value       x    c  c with   f  (      x    c    ,  c  )  =  0.Intuitively this seems true, but I woludn&#039;t know how to prove it.

kidoceanoe · Accepted Answer

The proposition is not true. Take   a  =  0 and   b  =  1, and let   f be given by  f  (  x  ,  y  )  =  1  −  2      x    2    y  f  (  x  ,  a  )  =  1  &amp;gt;  0 for all   x, and at       x    b    =  1 we have   f  (      x    b    ,  b  )  =  −  1  &amp;lt;  0, and   f is continuous.But if   c  =  0, we already know   f  (  x  ,  c  )  =  1 so there is no solution to   f  (      x    c    ,  c  )  =  0. If   c  &amp;gt;  0, then  f      (          c              −        1                  /                2              ,    c    )    =  −  1so it is not true that   f  (  x  ,  c  )  ≥  0 for all   x.

Let's consider the continuous function f:R*[a,b]->R Such that f(x,a)>0 for all x and there exists x_b with f(xb,b)<=0. Then there exists c in [a,b] such that f(x,c)>=0 for all x and a real value x_c with f(x_c,c)=0. Intuitively this seems true, but I woludn't know how to prove it.

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