Let f : [ a , b ] → R be a continuous function and let f ( a ) &lt; f ( b ). Then by intermediate value theorem ( 1 ) f ( [ a , b ] ) = [ f ( a ) , f ( b ) ] ( 2 ) f ( [ a , b ] ) ⊇ [ f ( a ) , f ( b ) ] ( 3 ) f ( [ a , b ] ) ⊆ [ f ( a ) , f ( b ) ] ( 4 ) f ( [ a , b ] ) ≠ [ f ( a ) , f ( b ) ]For example if I take the function f ( x ) = x 2 on [ − 1 , 2 ] then f ( [ − 1 , 2 ] ) = [ 0 , 4 ] which is not equal to [ f ( − 1 ) , f ( 2 ) ] = [ 1 , 4 ], also [ 1 , 4 ] ⊆ [ 0 , 4 ], so options 1 is incorrect. But how to use IVT here? Please help.

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Let   f  :  [  a  ,  b  ]  →      R   be a continuous function and let   f  (  a  )  &amp;lt;  f  (  b  ). Then by intermediate value theorem  (  1  )  f  (  [  a  ,  b  ]  )  =  [  f  (  a  )  ,  f  (  b  )  ]    (  2  )  f  (  [  a  ,  b  ]  )  ⊇  [  f  (  a  )  ,  f  (  b  )  ]    (  3  )  f  (  [  a  ,  b  ]  )  ⊆  [  f  (  a  )  ,  f  (  b  )  ]    (  4  )  f  (  [  a  ,  b  ]  )  ≠  [  f  (  a  )  ,  f  (  b  )  ]For example if I take the function   f  (  x  )  =      x    2   on   [  −  1  ,  2  ] then   f  (  [  −  1  ,  2  ]  )  =  [  0  ,  4  ] which is not equal to   [  f  (  −  1  )  ,  f  (  2  )  ]  =  [  1  ,  4  ], also   [  1  ,  4  ]  ⊆  [  0  ,  4  ], so options 1 is incorrect. But how to use IVT here? Please help.

ta1toerx68 · Accepted Answer

Let   c  ∈  [  f  (  a  )  ,  f  (  b  )  ] then by IVT (since   f is continuous) it follows that there exists some   x  ∈  [  a  ,  b  ] such that   f  (  x  )  =  c. Therefore   c  ∈  f  (  [  a  ,  b  ]  ) (the image of interval   [  a  ,  b  ] under   f). Thus  [  f  (  a  )  ,  f  (  b  )  ]  ⊆  f  (  [  a  ,  b  ]  )

gentyusag5jzt · Accepted Answer

IVT says that   f take all value between   f  (  a  ) and   f  (  b  ). In other word that  [  f  (  a  )  ,  f  (  b  )  ]  ⊆  f  (  [  a  ,  b  ]  )  .

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