let f : R → R be a continuous function such that f ( 0 ) ≤ 1 and for all real x, f ( x ) 2 − 3 f ( x ) + 2 ≥ 0. Prove that f ( x ) ≤ 1 for all real x.I think you would start by factorising to get ( f ( x ) − 2 ) ( f ( x ) − 1 ) ≥ 0 and then f ( x ) ≥ 1 but I'm not really sure where to go from there? I think maybe you can use the intermediate value theorem but I'm not exactly sure how. TIA

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let   f  :      R    →      R   be a continuous function such that   f  (  0  )  ≤  1 and for all real   x,   f  (  x      )    2    −  3  f  (  x  )  +  2  ≥  0. Prove that   f  (  x  )  ≤  1 for all real   x.I think you would start by factorising to get   (  f  (  x  )  −  2  )  (  f  (  x  )  −  1  )  ≥  0 and then   f  (  x  )  ≥  1 but I&#039;m not really sure where to go from there? I think maybe you can use the intermediate value theorem but I&#039;m not exactly sure how. TIA

Aliya Moore · Accepted Answer

Factorizing is a good start. Suppose now, there were some   x with   f  (  x  )  &amp;gt;  1. By the intermediate value theorem due to   f  (  0  )  ≤  1, there is some       x    0   with   1  &amp;lt;  f  (      x    0    )  &amp;lt;  2. But then            (        f  (      x    0    )  −  2            )                  (        f  (      x    0    )  −  1            )        &amp;lt;  0Contradiction.

let f:R->R be a continuous function such that f(0)<=1 and for all real x, f(x)^2−3f(x)+2>=0. Prove that f(x)<=1 for all real x.

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