1. ## maximum value

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

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

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

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

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

$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

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

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

$5x - 8 = 0$

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

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

for $\frac{8}{5} < x < 5$ , 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 ...

$y(0) = 0$

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

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