Find the x value where the function has a horizontal tangent line.

[coooool222]

$$f(z) = 1/3 z^3 -6(1/2) z^2+36z+11$$$$f'(z)=z^2-6z+36$$
Now I derived the equation so now:
$$0=z^2-6z+36$$
How can I find the x value where the function has a horizontal tangent line if I can't factor it?

 

[JeffM]

$$f(z) = 1/3 z^3 -6(1/2) z^2+36z+11$$$$f'(z)=z^2-6z+36$$
Now I derived the equation so now:
$$0=z^2-6z+36$$
How can I find the x value where the function has a horizontal tangent line if I can't factor it?

The problem does not lie in that there is no obvious factoring. The Fundamental Theorem of Algebra guarantees that there is a factoring. Nor is the actual factoring that hard to find. The real problem is that no real value of z makes that derivative zero.

Two possibilities. One is that there is a mistake in the problem as given to you. The other is that you made a mistake in presenting the problem. For example, does the problem perhaps say MINUS 36z rather than plus 36z.

 

[Otis]

Find the x value … f(z) = 1/3*z^3 − 6(1/2)*z^2 + 36z + 11

Hi coooool222. I agree with JeffM; the given function has no horizontal tangent. Also, they've asked for a single x-value, but their independent variable is z, and it's a cubic polynomial (so there can't be a single horizontal tangent). Additionally, it's peculiar for them to have factored only the squared term's coefficient (why express 3 as 6×½).

How can I find [roots of polynomial z^2−6z+36] if I can't factor it?

Other methods available for solving quadratic equations include Completing the Square and the Quadratic Formula.

Even though the exercise is flawed, your approach is still good. Show that the first derivative's roots are Complex numbers, and then report that f has no horizontal tangent. Or, change the exercise as suggested by Jeff and find z-values where the new f has horizontal tangents. (It's a cubic polynomial, so if it has any turning points then there must be two of them because such cubic graphs are S-shaped.)

$\;$

 

[Dr.Peterson]

$$f(z) = 1/3 z^3 -6(1/2) z^2+36z+11$$$$f'(z)=z^2-6z+36$$
Now I derived the equation so now:
$$0=z^2-6z+36$$
How can I find the x value where the function has a horizontal tangent line if I can't factor it?

Could you show us (perhaps as an image) the entire problem as given to you? As others have said, it is an odd problem as given to you; I suspect what you showed us is just part of a larger problem, and/or that you have paraphrased it.