Find the number of real roots of the function
Inspiration for this problem: How many roots?#2
This problem is part of the set " Xenophobia "
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Since − 1 ≤ cos 2 0 1 7 ( x ) ≤ 1 , the roots of f(x) exists at − 1 ≤ lo g 2 0 1 7 ( x ) ≤ 1 2 0 1 7 1 ≤ x ≤ 2 0 1 7 .
Divide the interval into subintervals ( ( k − 1 ) π , k π ) for k = 1 , 2 , . . . , ⌊ π 2 0 1 7 ⌋ = 6 4 2 .
(Note: It's fine to start with 0 instead of 2 0 1 7 1 since there is no root at x ≤ 2 0 1 7 1 )
By Intermediate Value Theorem, there will be exactly one root inside each of these subintervals.
For the leftover interval ( 6 4 2 π , 2 0 1 7 ) , since cos ( 6 4 2 π ) = 1 , the trigonometric part of the function will be decreasing while the logarithmic part is still increasing. These two parts will intersect again one last time; therefore f(x) will have a root at this interval.
642 roots from the subintervals + 1 additional root from the leftover interval = 6 4 3 total roots