For example, to prove that the function x 8 + x − 1 = 0 has exactly two roots, we first prove that f has at most 2 real roots by using differentiation. f ′ ( x ) = 8 x 7 + 1 = 0, by using that the function has two roots and therefore it should have at least one point such that f ′ ( c ) = 0 and then we use the IVT to prove that it has exactly two roots.However, the derivative of equation x 4 − 6 x 2 − 8 x + 1 = 0 has 2 roots but still, I should try to prove that it has exactly two roots. If its derivative had one solution, then I could first prove that it has two roots at most as I did but it has two roots, so how should we approach the problem?

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For example, to prove that the function       x    8    +  x  −  1  =  0 has exactly two roots, we first prove that   f has at most 2 real roots by using differentiation.       f    ′    (  x  )  =  8      x    7    +  1  =  0, by using that the function has two roots and therefore it should have at least one point such that   f      ′    (  c  )  =  0 and then we use the IVT to prove that it has exactly two roots.However, the derivative of equation       x    4    −  6      x    2    −  8  x  +  1  =  0 has 2 roots but still, I should try to prove that it has exactly two roots. If its derivative had one solution, then I could first prove that it has two roots at most as I did but it has two roots, so how should we approach the problem?

Raelynn Johnson · Accepted Answer

The derivative   4      x    3    −  12  x  −  8 factors into   (  x  +  1      )    2    (  x  −  2  ). Thus   −  1 is a double root at which the derivative doesn&#039;t change sign. Thus       x    4    −  6      x    2    −  8  x  +  1 is monotonically decreasing for   x  &amp;lt;  2 and increasing for   x  &amp;gt;  2. It&#039;s negative at   x  =  2 and goes to   ∞ for   x  →  ±  ∞, so it has exactly two roots.

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