# Thread: finding local max of function via 1st derivative

1. ## finding local max of function via 1st derivative

y=x(1-x)^(1/3)

The derivative of the function turns out to be (1-x)^(1/3) - x/(3(1-x)^2/3)
Using the calculator I found out the max is 0.750, .472...But when I set the derivative equal to 0...and tried to factor I got x=-3/11, x=1...I understand how to find the mix and max, my question is how to factor this particular equation...

2. When you set the derivative equal to zero, you need to do the following algebra to get the critical point:

$0=(1-x)^{\frac{1}{3}}-\frac{x}{3(1-x)^{\frac{2}{3}}}$

$(1-x)^{\frac{1}{3}}=\frac{x}{3(1-x)^{\frac{2}{3}}}$

$1-x=\frac{x^3}{27(1-x)^2}$

$27(1-x)^3=x^3$

$3(1-x)=x$

$3=4x$

$x=.75$

3. Originally Posted by bcahmel
y=x(1-x)^(1/3)

The derivative of the function turns out to be (1-x)^(1/3) - x/(3(1-x)^2/3)
Using the calculator I found out the max is 0.750, .472...But when I set the derivative equal to 0...and tried to factor I got x=-3/11, x=1...I understand how to find the mix and max, my question is how to factor this particular equation...
Is...

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

... so that the maximim is for $x=\frac{3}{4}$ ...

Merry Christmas from Italy

$\chi$ $\sigma$

4. Hello, bcahmel!

$y\:=\:x(1-x)^{\frac{1}{3}}$

$\text{The derivative turns out to be: }\:y' \:=\: (1-x)^{\frac{1}{3}} - \dfrac{x}{3(1-x)^{\frac{2}{3}}}$

$\text{Using the calculator I found out the max is: }\:(0.75,\;0.47247\hdots)$

We have: . $(1-x)^{\frac{1}{3}} - \dfrac{x}{3(1-x)^{\frac{2}{3}}} \;=\;0$

Assuming $x \ne 1$, multiply through by $3(1-x)^{\frac{2}{3}}\!:$

. . $3(1-x) - x \;=\;0 \quad\Rightarrow\quad 3 - 3x - x \:=\:0$

. . $-4x \:=\:-3 \quad\Rightarrow\quad x \:=\:\dfrac{3}{4}$

5. thanks to everyone for the help! I appreciate it greatly