To find the zeros of the function I(x) = (12x^3/5) - (27x/5), we need to set the function equal to zero and solve for x.
12x^3/5 - 27x/5 = 0
Multiply both sides of the equation by 5 to remove the denominators:
12x^3 - 27x = 0
Factor out the common term x:
x(12x^2 - 27) = 0
Set each factor equal to zero:
x = 0
12x^2 - 27 = 0
To solve the quadratic equation, we can factor it further:
12x^2 - 27 = 0
3(4x^2 - 9) = 0
3(2x - 3)(2x + 3) = 0
Setting each factor equal to zero:
2x - 3 = 0 or 2x + 3 = 0
Solving each equation separately:
2x - 3 = 0 --> 2x = 3 --> x = 3/2
2x + 3 = 0 --> 2x = -3 --> x = -3/2
Therefore, the zeros of the function I(x) = (12x^3/5) - (27x/5) are x = 0, x = 3/2, and x = -3/2.
Find the zeros of I(x)=12x^3/5-27x/5
1 answer