Proof that a polynomial has a minimum in RI have to prove to following statement and I am having a really hard time here.There it is:Prove that the following polynomial has a minimum in R p ( x ) = x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0 I tried to make the following proof, but I am stuck:The polynomial can be shown like this: x 4 ( 1 + a 3 x + a 2 x 2 + a 1 x 3 + a 0 x 4 ) Let g ( x ) = ( 1 + a 3 x + a 2 x 2 + a 1 x 3 + a 0 x 4 ) So we can now write the polynomial like p ( x ) = x 4 g ( x )Since x , a ∈ R we can say that lim x → ∞ = lim x → − ∞ = 1 (from limit arithmetic).We know that it is always the case that x 4 &gt; 0, so the sign of p ( x ) = x 4 g ( x )is determined by g(x). For certain values, g ( x ) &lt; 0 so we can say that there is some a , b ∈ R such that g ( a ) &lt; 0 and g ( b ) &gt; 0.p(x) is continouos everywhere, certainly in [a,b], so from the Extreme Value Theorem, the function has a minimum in [a,b].There are two problems in my proof:I can't find any value in which g(x) is negative, only if there is some a which is negative. What if a is always positive?If the first problem is solved, I managed to prove the statement for some [a,b]. Is it just enough? It seems to me that it isn't.

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Proof that a polynomial has a minimum in RI have to prove to following statement and I am having a really hard time here.There it is:Prove that the following polynomial has a minimum in R  p  (  x  )  =      x    4    +      a    3        x    3    +      a    2        x    2    +      a    1    x  +      a    0  I tried to make the following proof, but I am stuck:The polynomial can be shown like this:      x    4        (    1    +                  a        3            x        +                  a        2                    x        2              +                  a        1                    x        3              +                  a        0                    x        4              )  Let  g  (  x  )  =      (    1    +                  a        3            x        +                  a        2                    x        2              +                  a        1                    x        3              +                  a        0                    x        4              )  So we can now write the polynomial like  p  (  x  )  =      x    4    g  (  x  )Since   x  ,  a  ∈      R   we can say that       lim          x      →      ∞        =      lim          x      →      −      ∞        =  1 (from limit arithmetic).We know that it is always the case that       x    4    &amp;gt;  0, so the sign of  p  (  x  )  =      x    4    g  (  x  )is determined by g(x). For certain values,   g  (  x  )  &amp;lt;  0 so we can say that there is some   a  ,  b  ∈      R   such that   g  (  a  )  &amp;lt;  0 and   g  (  b  )  &amp;gt;  0.p(x) is continouos everywhere, certainly in [a,b], so from the Extreme Value Theorem, the function has a minimum in [a,b].There are two problems in my proof:I can&#039;t find any value in which g(x) is negative, only if there is some a which is negative. What if a is always positive?If the first problem is solved, I managed to prove the statement for some [a,b]. Is it just enough? It seems to me that it isn&#039;t.

pararevisarii · Accepted Answer

Step 1The claim itself is true more generally for any polynomial of even degree with positive leading coefficient.Step 2We do not need that   g  (  x  )  &amp;lt;  0 for some x and in fact such x may not exist. What we need is that       lim          x      →      ±      ∞        g  (  x  )  =  1 implies that there exists   M  ∈      R   such that       |    x      |    &amp;gt;  M implies       |    g  (  x  )  −  1      |    &amp;lt;      1    2   (so that   g  (  x  )  &amp;gt;      1    2  ). Then for       |    x      |    &amp;gt;  M we conclude   f  (  x  )  &amp;gt;      1    2        x    4    &amp;gt;      1    2        M    4  . We may assume wlog. that       1    2        M    4    &amp;gt;      a    0    =  f  (  0  ). Then the minimum the continuous function has on [-M,M] is in fact a global minimum.

Chaim Ferguson · Accepted Answer

Step 1You can&#039;t just write   f  (  x  )  =      x    4    g  (  x  ) because f(0) is defined and g(0) isn&#039;t. Instead, you can simply prove      lim          x      →      ±      ∞        f  (  x  )  =  ∞(because the       x    4   term dominates all others in the limit)With that settled, by the definition of these limits you can find some   R  &amp;gt;  0 such that  f  (  x  )  &amp;gt;      a    0      ∀  x  :      |    x      |    &amp;gt;  R(the   &amp;gt;      a    0   is arbitrary here, any constant works as long as f attains it inside [-R,R]; also, R need not be the optimal choice) Now look at   f            |              [      −      R      ,      R      ]      . This is a continuous function on a closed and bounded set. You should have a theorem giving you that   f            |              [      −      R      ,      R      ]       attains a minimum.Since   0  ∈  [  −  R  ,  R  ], we know that this minimum is at most   f  (  0  )  =      a    0  , so f doesn&#039;t attain smaller values outside of [-R,R] by construction of R.Step 2This proves      min          x      ∈              R              f  (  x  )  =      min          x      ∈      [      −      R      ,      R      ]        f  (  x  )  ∈      R  exists.

Prove that the following polynomial has a minimum in R p(x)=x^4+a_3 x^3+a_2 x^2+a_1x+a_0.

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