1. local extremes

Find function $\displaystyle y=\sqrt[3]{x^3+3x^2}$ local extremes.

I start:

$\displaystyle y'= \frac{1}{3}(x^3+3x^2)^{\frac{-2}{3}}*(x^3+3x^2)'=\frac{1}{3}(x^3+3x^2)^{\frac{-2}{3}}(3x^2+6x)$

$\displaystyle y'= \frac{1}{3(\sqrt[3]{(x^3+3x^2)^2})}(3x^2+6x)=0$

2. Originally Posted by Bernice
Find function $\displaystyle y=\sqrt[3]{x^3+3x^2}$ local extremes.

I start:

$\displaystyle y'= \frac{1}{3}(x^3+3x^2)^{\frac{-2}{3}}*(x^3+3x^2)'=\frac{1}{3}(x^3+3x^2)^{\frac{-2}{3}}(3x^2+6x)$

$\displaystyle y'= \frac{1}{3(\sqrt[3]{(x^3+3x^2)^2})}(3x^2+6x)=0$
now find the critical values of y' the root of the denominator and the numerator

$\displaystyle y'= \frac{1}{3(\sqrt[3]{(x^2(x+3))^2})}(3x(x+2))=0$

after you find the critical points study y' signs around critical points

3. Easier to consider the function

$\displaystyle f(x)=x^3+3x^2$
$\displaystyle f'(x)=3x^2+6x$
$\displaystyle f'(x)=3x(x+2)$
f'(x)=0 at x=0 and x=-2
f(x) has its local extremes at x=0 and x=-2.

Hence, y has its local extremes at x=0 (minima) and x=-2 (maxima).

[Use 2nd derivative test to figure if it's a minima or maxima.
f''(x)=6(x+1)
f''(0)=6>0 (indicates minima)
f''(-2)=-6<0 (indicates maxima)
]

The local extremes are y=0 (minima) and $\displaystyle \sqrt[3]{4}$ (maxima).

4. Originally Posted by alexmahone
Easier to consider the function

$\displaystyle f(x)=x^3+3x^2$
$\displaystyle f'(x)=3x^2+6x$
$\displaystyle f'(x)=3x(x+2)$
$\displaystyle f'(x)=0 at x=0 and x=-2$
$\displaystyle f(x) has its local extremes (minima) at x=0 and x=-2$
what you said is true for this function but if the power of the denominator is odd number (always positive) you can't use you method