We are finding critical numbers
f(x)= 8x^3+x^2+8x
f'(x)=24x^2+2x+8 -> 2(12x^2+x+4)
I'm honestly stuck. All I can tell is that I need to set my f'(x)=0, but I am lost there too.
Quadratic form. equates to (-1±i√191)/24 I believe. That is as far as I can get.
To find the critical numbers of a function, you start by finding the derivative of the function and then setting it equal to zero. In this case, you have correctly found the derivative of f(x) as f'(x) = 24x^2 + 2x + 8.
To solve the equation f'(x) = 0, you can use the quadratic formula. The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In your case, the equation is 24x^2 + 2x + 8 = 0 and you can identify a, b, and c as follows:
a = 24
b = 2
c = 8
Plugging these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4*24*8)) / (2*24)
x = (-2 ± √(4 - 768)) / 48
x = (-2 ± √(-764)) / 48
At this point, we see that the expression inside the square root (√(-764)) is negative. This indicates that the quadratic equation has no real solutions. Instead, it has complex solutions. Specifically, the solutions can be expressed as (-1 ± i√191) / 24, as you correctly stated.
Therefore, in this particular case, the critical numbers of the function f(x) = 8x^3 + x^2 + 8x are (-1 + i√191) / 24 and (-1 - i√191) / 24.