Does #x^3+sin x = 1# have a root in #(0, 2pi)# ?
1 Answer
Feb 5, 2018
Yes, it has one root in
Explanation:
Let
Note that:
-
#f(0) = 0+0-1 = -1 < 0# -
#f(pi/2) = pi^3/8+1-1 = pi^3/8 > 0# -
Both
#x^3# and#sin x# are continuous and strictly monotonically increasing in#(0, pi/2]# , so#f(x)# is too. -
#pi/2 ~~ 1.57 > root(3)(2)# , so when#x >= pi/2# ,#f(x) > 2+sin(x)-1 > 0#
Hence by Bolzano's theorem
graph{x^3+sinx -1 [-7, 7, -20, 20]}