# Finding crritical points of a piecewise function

• Jun 11th 2009, 08:34 PM
jimmyp
Finding crritical points of a piecewise function
Given:

$\displaystyle f(x)=\left\{\begin{array}{cc}-x^3+1 ,&\mbox{ if } x\leq 0\\-x^2 +2x, & \mbox{ if } x>0\end{array}\right.$

Find the critical numbers and open intervals on which the function is increasing or decreasing.

Would I simply compute the derivative of -x^3+1 and -x^2+2x seperately and solve each down?

Can anyone offer some instruction for piecewise functions ?
• Jun 11th 2009, 09:27 PM
VonNemo19
Quote:

Originally Posted by jimmyp
Given:

$\displaystyle f(x)=\left\{\begin{array}{cc}-x^3+1 ,&\mbox{ if } x\leq 0\\-x^2 +2x, & \mbox{ if } x>0\end{array}\right.$

Find the critical numbers and open intervals on which the function is increasing or decreasing.

Would I simply compute the derivative of -x^3+1 and -x^2+2x seperately and solve each down?

Can anyone offer some instruction for piecewise functions ?

yes

all youre trying to do is find the same things you would in a regular function.

sketch the graph and then estimate where the critical values are. see if this is a good approximation of doing it analytically
• Jun 12th 2009, 03:01 PM
jimmyp
Thus:

y' = $\displaystyle -3x^2$
and
$\displaystyle -2x+2$

$\displaystyle -3x^2=0$
$\displaystyle x=0$

and:

$\displaystyle -2x+2=0$
$\displaystyle x=1$

Critical point: $\displaystyle (0,1)$

And the test intervals are: $\displaystyle (- \infty, 0),(1, \infty)$

Thus:

$\displaystyle - \infty < x < 0$
test value: -2
Sign of f'(x) on -2x+2 = -2
Sign of f'(x) on -3x^2 = -12
Conclusion: decreasing for x < 0 and decreasing for x > 0

and:

$\displaystyle \infty < x < 1$
test value: 2
Sign of f'(x) on -2x+2 = -2
Sign of f'(x) on -3x^2 = -12
Conclusion: decreasing for x < 0 and decreasing for x > 0

is this correct or have i made a mistake ?
• Jun 12th 2009, 06:16 PM
pankaj
The intervals of increasing monotonicity are $\displaystyle (0,1)$ and the interval where $\displaystyle f(x)$ decreases are $\displaystyle (-\infty,0)\cup (1,\infty)$

You forgot to check for the sign of derivative on the interval (0,1)