Prove that [tex](x^5)-(x^2)+2x +3=0[/tex] has at least one real root.

Prove that [tex]x = \cos(x)[/tex] has at least one solution.

The first one is of order 5, so it has either 1, 3 or 5 real roots (unless any coefficent was complex). Proof complete :) The other one, if it has a solution, it must be in [-1;1]. Because it only gives positive results the solution is further restricted to [0;1]. Because the cosine function is continuous and strictly decreasing on this interval, the difference of x and it's cosine will shrink up to some point within the interval where it gets to 0 (the solution) and then flips sign (the cosine gets less than the number), further decreasing until the end of the interval.

Answered by Anonymous | 2024-06-10

ok no cap ;

Answered by wilson007what | 2024-06-12

The polynomial x 5 − x 2 + 2 x + 3 = 0 has at least one real root due to the Intermediate Value Theorem, as it changes signs between evaluated points. Similarly, the equation x = cos ( x ) has at least one solution because both x and cos ( x ) intersect within the interval [ − 1 , 1 ] . Thus, both equations have at least one real root or solution.
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Answered by Anonymous | 2025-06-15