Let f ( x ) = 20 ( x 6 + x 4 + x 2 + 1 ) .I need to show that for any k ∈ [ 5 , 20 ], there is a point c ∈ [ 0 , 1 ] such that f ( c ) = k.Thanks in advance.

Question

Let   f  (  x  )  =            20              (                  x          6                +                  x          4                +                  x          2                +        1        )            .I need to show that for any   k  ∈  [  5  ,  20  ], there is a point   c  ∈  [  0  ,  1  ] such that   f  (  c  )  =  k.Thanks in advance.

prelatiuvq · Accepted Answer

Observe that       x    6    +      x    4    +      x    2    +  1  &amp;gt;  0    ∀  x  ∈      R   and is continuous. Now   f  (  x  ) is continuous on       R   and observe that   f  (  1  )  =  5  ,  f  (  0  )  =  20. Intermediate value theorem tells that for every   k such that   5  ≤  k  ≤  20 there exists a   c  ∈  [  0  ,  1  ] such that   f  (  c  )  =  k.

Let f(x)=20/(x^6+x^4+x^2+1). I need to show that for any k in [5,20], there is a point c in [0,1] such that f(c)=k.

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