Suppose f : [ 1 , 2 ] → [ 5 , 7 ] is continuous. Show that f ( c ) = 2 c + 3 for some c ∈ [ 1 , 2 ].First note f ( 1 ) = 5 and f ( 2 ) = 7. By the IVT, all values c ∈ [ 1 , 2 ] are hit. I'm just wondering how to put all of these facts together to arrive at f(c)=2c+3 for all c ∈ [ 1 , 2 ].

Question

Suppose   f  :  [  1  ,  2  ]  →  [  5  ,  7  ] is continuous. Show that   f  (  c  )  =  2  c  +  3 for some   c  ∈  [  1  ,  2  ].First note   f  (  1  )  =  5 and   f  (  2  )  =  7. By the IVT, all values   c  ∈  [  1  ,  2  ] are hit. I&#039;m just wondering how to put all of these facts together to arrive at f(c)=2c+3 for all   c  ∈  [  1  ,  2  ].

Audrey Rosales · Accepted Answer

Look at the function  g  (  x  )  =  f  (  x  )  −  2  x  −  3We have   g  (  1  )  =  f  (  1  )  −  5  ≥  0 and   g  (  2  )  =  f  (  2  )  −  7  ≤  0 since   f  (  x  )  ∈  [  5  ,  7  ] for all   x  ∈  [  1  ,  2  ].Hence, by intermediate value theorem, there is a   c  ∈  [  1  ,  2  ] such that   g  (  c  )  =  0. Hence, there exists a   c  ∈  [  1  ,  2  ] such that  f  (  c  )  =  2  c  +  3

Suppose f:[1,2]->[5,7] is continuous. Show that f(c)=2c+3 for some c in [1,2]. First note f(1)=5 and f(2)=7. By the IVT, all values c in [1,2] are hit. I'm just wondering how to put all of these facts together to arrive at f(c)=2c+3 for all c in [1,2].

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