• Nov 29th 2009, 03:12 PM
jhonwashington
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
• Nov 29th 2009, 03:30 PM
Bacterius
When the derivative of the function in a point is $0$, its tangent is horizontal.

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

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

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

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

$f'(x) = 0$

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

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

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

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

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

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

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

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