Given equation x 100 + x 99 + x 98 + . . . + x + 1 = 5 I need to determine whether there exists one solution x such that x &gt; 0. Also, whether I have one solution x &lt; 0.First, I let a = 0 such that f ( a ) = 1. Let b = 1 such that f ( b ) = 101. Since f ( a ) &lt; 5 &lt; f ( b ), thus, by intermediate value theorem, there exits one solution x ∈ ( 0 , 1 ). Similarly, I can conclude that there exits one solution smaller than 0. Is my process correct?

Question

Given equation       x          100        +      x          99        +      x          98        +  .  .  .  +  x  +  1  =  5 I need to determine whether there exists one solution   x such that   x  &amp;gt;  0. Also, whether I have one solution   x  &amp;lt;  0.First, I let   a  =  0 such that   f  (  a  )  =  1. Let   b  =  1 such that   f  (  b  )  =  101. Since   f  (  a  )  &amp;lt;  5  &amp;lt;  f  (  b  ), thus, by intermediate value theorem, there exits one solution   x  ∈  (  0  ,  1  ). Similarly, I can conclude that there exits one solution smaller than 0. Is my process correct?

Ari Jacobs · Accepted Answer

Yes since the function is continuous and, as noticed in the comments, by geometric series we have for   x  ≠  1  f  (  x  )  =                    x                  101                    −      1              x      −      1      therefore  f  (  −  1  )  =  1    f  (  −  2  )  &amp;gt;  5also for   x  &amp;lt;  0 at least a solution exists in the interval   (  −  2  ,  −  1  ).

Given equation x <mrow class="MJX-TeXAtom-ORD"> 100 </mrow> </msup>

Answered question

Answer & Explanation

New Questions in Calculus 2