# show function has stationary points

• February 24th 2010, 05:40 AM
decoy808
show function has stationary points
need help with this one.

show the function below has stationary points @ x=-1 x=0 x= 2

$y=\frac{x^4}{4}-\frac{x^3}{3}-x^2$

$12y=3x^4-4x^3-12x^2$

$
12\frac{dy}{dx}=12x^3-12x^2-24x
$

$
\frac{dy}{dx}=x^3-x^2-2x
$

$
\frac{dy}{dx}=0
$

correct so far?
• February 24th 2010, 05:49 AM
Quote:

Originally Posted by decoy808
need help with this one.

show the function below has stationary points @ x=-1 x=0 x= 2

$y=\frac{x^4}{4}-\frac{x^3}{3}-x^2$

$12y=3x^4-4x^3-12x^2$

$
12\frac{dy}{dx}=12x^3-12x^2-24x
$

$
\frac{dy}{dx}=x^3-x^2-2x
$

$
\frac{dy}{dx}=0
$

correct so far?

yes . Then carry on with the factorization .
• February 24th 2010, 08:50 AM
HallsofIvy
Quote:

Originally Posted by decoy808
need help with this one.

show the function below has stationary points @ x=-1 x=0 x= 2

$y=\frac{x^4}{4}-\frac{x^3}{3}-x^2$

$12y=3x^4-4x^3-12x^2$

$
12\frac{dy}{dx}=12x^3-12x^2-24x
$

$
\frac{dy}{dx}=x^3-x^2-2x
$

$
\frac{dy}{dx}=0
$

correct so far?

Yes, so solve $x^3- x^2- 2x= 0$ or, since you are only asked to show that the stationary points are at x= -1, x= 0, and x= 2, evaluate $x^3- x^2- 2x$ at each of those values of x.