# show function has stationary points

• Feb 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

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

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

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

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

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

correct so far?
• Feb 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

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

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

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

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

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

correct so far?

yes . Then carry on with the factorization .
• Feb 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

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

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

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

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

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

correct so far?

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