Let f ( x ) = x 6 − 1 3 x − 1 Prove that the range of f is R .( Hint: use the Intermediate Value Theorem.)I thought IVT was meant to show that the function has a root? Please help, I don't know how I can use IVT to prove the range.

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Let  f  (  x  )  =                    x        6            −      1              3      x      −      1      Prove that the range of   f is       R  .( Hint: use the Intermediate Value Theorem.)I thought IVT was meant to show that the function has a root? Please help, I don&#039;t know how I can use IVT to prove the range.

Dobermann82 · Accepted Answer

You have       lim          x      →      −      ∞        f  (  x  )  =  −  ∞ and       lim          x      →                        1          3                −              f  (  x  )  =  ∞.Moreover   f is continuous on the interval   (  −  ∞  ,      1    3    ). Therefore by the IVT, the image of   (  −  ∞  ,      1    3    ) under   f is equal to       R  . A fortiori, the image of   f is equal to       R  .

Wade Bullock · Accepted Answer

Hint: If   y  ∈      R  , then asserting that   y belongs to the range of   f is the same thing as asserting that the equation   f  (  x  )  −  y  =  0 has a root.

Let f ( x ) = x 6 </msup> &#x22

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