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, we need to find the values of x for which the derivative of the function is equal to zero or undefined. In this case, let's start by finding the values where the derivative is equal to zero.
You correctly found the derivative of the function:
f'(x) = 24x^2 + 2x + 8
To find the values of x for which the derivative is equal to zero, we need to solve the equation f'(x) = 0.
Setting the derivative equal to zero:
24x^2 + 2x + 8 = 0
Now, we have a quadratic equation. To solve this, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 24, b = 2, and c = 8. Let's substitute these values into the formula:
x = (-2 ± √(2^2 - 4 * 24 * 8)) / (2 * 24)
x = (-2 ± √(4 - 768)) / 48
x = (-2 ± √(-764)) / 48
Since we have a negative value inside the square root, we know that there are no real solutions. This means that there are no critical numbers in this case.
However, it's worth noting that the values obtained from the quadratic formula, x = (-1 ± i√191) / 24, are complex numbers. While complex numbers are not considered as critical numbers for real-valued functions, they can have significance in some contexts.
Therefore, in this case, there are no real critical numbers, but there are complex solutions obtained from the quadratic formula.