1. ## maximum value

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

What is the maximum value that this function takes on the interval [0,5]. (not plug and chug)

2. Originally Posted by genlovesmusic09
$\displaystyle y = {x^{\frac{2} {3}}}(x - 4)$

What is the maximum value that this function takes on the interval [0,5]. (not plug and chug)
do a function analysis using the first derivative test to look for extrema on the given interval.

3. Originally Posted by skeeter
do a function analysis using the first derivative test to look for extrema on the given interval.
I found the derivative to be $\displaystyle \frac{2} {3}{x^{\frac{{ - 1}} {3}}}(x - 4) + {x^{\frac{2} {3}}}(1)$
i then set it equal to zero and found x=1.6
and plugged 1.6 into the original equation and found the minimum of (about) -3.283

4. Originally Posted by genlovesmusic09
I found the derivative to be $\displaystyle \frac{2} {3}{x^{\frac{{ - 1}} {3}}}(x - 4) + {x^{\frac{2} {3}}}(1)$
i then set it equal to zero and found x=1.6
and plugged 1.6 into the original equation and found the minimum of (about) -3.283
however ...

What is the maximum value that this function takes on the interval [0,5].

5. i know that is where i am lost

6. Originally Posted by genlovesmusic09
$\displaystyle y = {x^{\frac{2} {3}}}(x - 4)$

What is the maximum value that this function takes on the interval [0,5]. (not plug and chug)
$\displaystyle y = x^{\frac{5}{3}} - 4x^{\frac{2}{3}}$

$\displaystyle y' = \frac{5}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}}$

note that x = 0 is a critical value

set y' = 0

$\displaystyle \frac{5}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}} = 0$

multiply every term by $\displaystyle 3x^{\frac{1}{3}}$ ...

$\displaystyle 5x - 8 = 0$

$\displaystyle x = \frac{8}{5}$

for $\displaystyle 0 < x < \frac{8}{5}$ , y'(x) < 0 ... y is decreasing

for $\displaystyle \frac{8}{5} < x < 5$ , y'(x) > 0 ... y is increasing

since the function decreases, then increases, that says the function has maxima at the endpoints. comparing the function values at the endpoints ...

$\displaystyle y(0) = 0$

$\displaystyle y(5) = 5^{\frac{2}{3}} > 0$

$\displaystyle y(5) = 5^{\frac{2}{3}}$ is the maximum over the interval.