How do you solve #x^3-2x>1#?

1 Answer
Sep 9, 2017

Answer:

#(-1, (1 - sqrt5)/2) and ((1 + sqrt5)/2, +inf)#

Explanation:

#f(x) = x^3 - 2x - 1 > 0#
The simplest method is graphing the function f(x), then looking for the parts of the graph that are above the x-axis.
graph{x^3 - 2x - 1 [-2.5, 2.5, -1.25, 1.25]}
On this graph, the x-intercepts are approximately:
x = - 1 , x = - 0.65, and x = 1.60
Answers by intervals:
(-1, - 0.65) and (1.60, + inf.)

Note 1 . We can find the exact values of the 3 x-intercepts by solving the equation
f(x) = x^3 - 2x - 1 = (x + 1)(x^2 - x - 1) = 0
The quadratic equation (x^2 - x - 1) = 0 gives 2 real roots:
x = (1 +- sqrt5)/2.
Therefor, the answers are:
#(-1, (1 - sqrt5)/2)# and #((1 + sqrt5)/2 , +inf.)#
Note 2 . Graphing calculator may give much more accurate values of the three x-intercepts.
Note 3. Another method is solving algebraically the inequality by creating a Sign Chart. See algebra books.