Finding crritical points of a piecewise function

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June 11th 2009, 08:34 PM
jimmyp

Finding crritical points of a piecewise function

Given:

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

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 ?
June 11th 2009, 09:27 PM
VonNemo19

Quote:

Originally Posted by jimmyp

Given:

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

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
June 12th 2009, 03:01 PM
jimmyp

Thus:

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

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

and:

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

Critical point: $(0,1)$

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

Thus:

$- \infty < x < 0$
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:

$\infty < x < 1$
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 ?
June 12th 2009, 06:16 PM
pankaj

The intervals of increasing monotonicity are $(0,1)$ and the interval where $f(x)$ decreases are $(-\infty,0)\cup (1,\infty)$

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