Solve for x, x ϵ R
¨Cx3 + 25x ¡Ü 0
To solve the inequality ¨Cx^3 + 25x ≤ 0, we need to find the values of x that satisfy the inequality.
Step 1: Start by factoring out the common factor, which in this case is x:
x(x^2 + 25) ≤ 0
Step 2: Determine the critical points where the inequality could change. In this case, we have x = 0 and x^2 + 25 = 0.
Setting x = 0, we get:
0(x^2 + 25) ≤ 0
0 ≤ 0 (which is always true)
Setting x^2 + 25 = 0, we get:
x^2 = -25
Since the square of any real number cannot be negative, this equation has no real solutions.
Step 3: Analyze the intervals:
We have two critical points: x = 0 and x^2 + 25 = 0.
For x < 0:
Plug in a test value, let's say x = -1:
-1(1 + 25) ≤ 0
-26 ≤ 0 (true)
For x > 0:
Plug in a test value, let's say x = 1:
1(1 + 25) ≤ 0
26 ≤ 0 (false)
For x = 0:
0(0 + 25) ≤ 0
0 ≤ 0 (true)
Step 4: Determine the solution:
The solution to the inequality ¨Cx^3 + 25x ≤ 0 is:
x ≤ 0
Therefore, the values of x that satisfy the inequality are all real numbers less than or equal to zero: x ϵ (-∞, 0].