Find the coordinates of points where the graph f(x) has horizontal tangents. As a check, graph f(x) and see whether the points you find look as though they have horizontal tangents

F(x)= 1/3x^3-3x^2-16x+8

f(x)=-141.3
f(x)=25.3
is that correct?
thanks

2. When the derivative of the function in a point is $\displaystyle 0$, its tangent is horizontal.

Your function is $\displaystyle f(x) = \frac{1}{3} x^3 - 3x^2 - 16x + 8$

Take the derivative of this function (using the sum and power differentiation principles) :

$\displaystyle f'(x) = x^2 - 6x - 16$

Now you are looking for x-coordinates when the tangent is horizontal. Thus, you want to solve $\displaystyle f'(x) = 0$ :

$\displaystyle f'(x) = 0$

$\displaystyle x^2 - 6x - 16 = 0$

It factorizes ! $\displaystyle -6 = 2 - 8$, $\displaystyle 16 = 2 \times (-8)$ :

$\displaystyle (x - 8)(x + 2) = 0$

Thus, $\displaystyle f(x)$ has an horizontal tangent in x-coordinates $\displaystyle 8$ and $\displaystyle -2$. But you also want y-coordinates, right ? So plug this back into your first equation :

$\displaystyle f(8) = \frac{-424}{3}$

$\displaystyle f(-2) = \frac{76}{3}$

Thus, the curve has horizontal tangents in coordinates $\displaystyle (8, f(8))$, and $\displaystyle (-2, f(-2))$, that is, $\displaystyle (8, \frac{-424}{3})$ and $\displaystyle (-2, \frac{76}{3})$.

So you are right (prefer keeping fractional answer though).