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.
so, what's the trouble? You have critical numbers where f'=0. But, as you found out, f' is never zero.
So, ... there are no critical numbers. There is no place where the tangent line is horizontal. Just look at the graph and sigh:
http://www.wolframalpha.com/input/?i=8x^3%2Bx^2%2B8x
To find the critical numbers of the function f(x) = 8x^3 + x^2 + 8x, you correctly started by finding the derivative, which is given by f'(x) = 2(12x^2 + x + 4).
To find the critical numbers, we need to solve the equation f'(x) = 0. In this case, we have:
2(12x^2 + x + 4) = 0
To solve this equation, let's first solve the quadratic expression inside the parentheses:
12x^2 + x + 4 = 0
Next, we can apply the quadratic formula, which states that for any quadratic equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 12, b = 1, and c = 4. Plugging these values into the quadratic formula, we get:
x = (-1 ± √(1^2 - 4 * 12 * 4)) / (2 * 12)
Simplifying further:
x = (-1 ± √(1 - 192)) / 24
x = (-1 ± √(-191)) / 24
Here, we encounter a complex number since the expression inside the square root is negative. So, the critical numbers in this case are complex numbers and can be written as:
x = (-1 ± i√191) / 24
Therefore, the critical numbers of the function f(x) = 8x^3 + x^2 + 8x are given by (-1 + i√191)/24 and (-1 - i√191)/24.
Remember that critical numbers represent the x-values where the derivative is equal to zero or where it is undefined, which can indicate local maxima, local minima, or points of inflection on the graph of the function.