Find the zeros of I(x)=12x^3/5-27x/5
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.
To find the zeros of the function I(x) = 12x^(3/5) - 27x/5, we set the function equal to zero and solve for x.
12x^(3/5) - 27x/5 = 0
To make the equation more manageable, we can multiply the entire equation by 5 to clear the fraction:
5(12x^(3/5)) - 5(27x/5) = 0
60x^(3/5) - 27x = 0
Now, we can factor out an "x" from both terms:
x(60x^(2/5) - 27) = 0
From this equation, we can see that x = 0 is one of the zeros.
To find the other zero, we set (60x^(2/5) - 27) equal to zero:
60x^(2/5) - 27 = 0
Adding 27 to both sides:
60x^(2/5) = 27
Dividing both sides by 60:
x^(2/5) = 27/60
Simplifying:
x^(2/5) = 9/20
Now, we raise both sides of the equation to the power of 5 to eliminate the fractional exponent:
(x^(2/5))^5 = (9/20)^5
x^2 = (9/20)^5
Taking the square root of both sides:
x = ±√((9/20)^5)
You can use a calculator to approximate the value of (√((9/20)^5)) which will give you the second zero of the function I(x).