Let f ( x ) = x 5 − 5 x + p. Show that f can have at most one root in [ 0 , 1 ],regardless of the value of p.This seems to be an IVT problem, so I will go forth with that: f ( a ) &gt; u &gt; f ( b ) f ( 0 ) &gt; 0 &gt; f ( 1 ) p &gt; 0 &gt; p − 4This is only true for p : ( 0 , 4 ) ie 0 &lt; p &lt; 4Now I have proved that there is atleast one root between [ 0 , 1 ], now I need to prove that there isn't a second.Would Rolles theorem be what I need here, or is there a simple deduction I can make from above to finish the problem?

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Let   f  (  x  )  =      x    5    −  5  x  +  p. Show that   f can have at most one root in   [  0  ,  1  ],regardless of the value of p.This seems to be an IVT problem, so I will go forth with that:  f  (  a  )  &amp;gt;  u  &amp;gt;  f  (  b  )  f  (  0  )  &amp;gt;  0  &amp;gt;  f  (  1  )  p  &amp;gt;  0  &amp;gt;  p  −  4This is only true for   p  :  (  0  ,  4  ) ie   0  &amp;lt;  p  &amp;lt;  4Now I have proved that there is atleast one root between   [  0  ,  1  ], now I need to prove that there isn&#039;t a second.Would Rolles theorem be what I need here, or is there a simple deduction I can make from above to finish the problem?

Hamza Conrad · Accepted Answer

Yes, Rolle&#039;s theorem is perhaps the best tool for this. If   f had two zeros, then the derivative  f  (  x  )  =  5      x    4    −  5  =  5  (      x    4    −  1  )would vanish at a point in the interior of   [  0  ,  1  ] - however, this isn&#039;t possible since       x    4    −  1 is negative for all such   x.

moiraudjpdn · Accepted Answer

First you can see that       f    ′    (  x  )  =  5  (      x    4    −  1  ) is negative for the given interval. Therefore, the given function is a decreasing function. So, depending upon the value of the function at 0 and 1 it will have one or no roots. Hope this helps!

Let f(x)=x^5−>5x+p. Show that f can have at most one root in [0,1],regardless of the value of p.

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