Note: I know how to prove this for f ( a ) = f ( a + 100 ), but no clue how to do it for the value that's asked of me.There's a continuous function f ( x ) : R → R With these properties:1. f ( x ) ≤ 2 for x ≥ 2002. f ( x ) ≤ 1 for x ≤ 1003. f ( x ) = 3 for x = 150I'm supposed to show that there exists an a for which the property f ( a ) = ( a + 200 ) is true. I've tried applying the intermediate value theorem every which way, but at this point I don't know if that's the right way to go.

Question

Note: I know how to prove this for   f  (  a  )  =  f  (  a  +  100  ), but no clue how to do it for the value that&#039;s asked of me.There&#039;s a continuous function   f  (  x  )  :      R    →         R  With these properties:1.   f  (  x  )  ≤  2     for     x  ≥  2002.   f  (  x  )  ≤  1     for     x  ≤  1003.   f  (  x  )  =  3     for     x  =  150I&#039;m supposed to show that there exists an   a for which the property   f  (  a  )  =  (  a  +  200  ) is true. I&#039;ve tried applying the intermediate value theorem every which way, but at this point I don&#039;t know if that&#039;s the right way to go.

Gemma Conley · Accepted Answer

Let   g  (  y  )  =  f  (  y  +  200  )  −  f  (  y  ). Then   g is continuous. We have   g  (  −  50  )  =  f  (  150  )  −  f  (  −  50  )  =  3  −  f  (  −  50  )  ≥  3  −  1  =  2  &amp;gt;  0. On the other hand,   g  (  150  )  =  f  (  350  )  −  f  (  150  )  ≤  2  −  3  =  −  1  &amp;lt;  0. Hence, there is a point   a between −50 and 150 where   g  (  a  )  =  0.

I know how to prove this for f(a)=f(a+100), but no clue how to do it for the value that's asked of me. There's a continuous function f(x):RR-> RR With these properties: f(x)<=2 for x>=200 f(x)<=1 for x≤100 f(x)=3 for x=150

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