# How do you find the critical numbers for h(t) = (t^(3/4)) − (9*t^(1/4)) to determine the maximum and minimum?

$\left(9 , - 2 \cdot {9}^{0.75}\right)$

#### Explanation:

Differentiate $h \left(t\right)$ with respect to t.

$h \left(t\right) = {t}^{\frac{3}{4}} - 9 \cdot {t}^{\frac{1}{4}}$

$h ' \left(t\right) = \left(\frac{3}{4}\right) \cdot {t}^{\frac{3}{4} - 1} - 9 \left(\frac{1}{4}\right) {t}^{\frac{1}{4} - 1}$

$h ' \left(t\right) = \left(\frac{3}{4}\right) \cdot {t}^{- \frac{1}{4}} - \left(\frac{9}{4}\right) {t}^{- \frac{3}{4}}$

After finding the derivative $h ' \left(t\right)$
Set $h ' \left(t\right) = 0$
$\left(\frac{3}{4}\right) \cdot {t}^{- \frac{1}{4}} - \left(\frac{9}{4}\right) {t}^{- \frac{3}{4}} = 0$
Multiply both sides by ${t}^{\frac{3}{4}}$
${t}^{\frac{3}{4}} \left(\left(\frac{3}{4}\right) \cdot {t}^{- \frac{1}{4}} - \left(\frac{9}{4}\right) {t}^{- \frac{3}{4}}\right) = 0 \cdot {t}^{\frac{3}{4}}$

$\left(\frac{3}{4}\right) \cdot {t}^{\frac{1}{2}} - \left(\frac{9}{4}\right) {t}^{0} = 0$
$\left(\frac{3}{4}\right) \cdot {t}^{\frac{1}{2}} - \frac{9}{4} = 0$
$\left(\frac{3}{4}\right) \cdot {t}^{\frac{1}{2}} = \frac{9}{4}$

${t}^{\frac{1}{2}} = 3$

$t = 9$

Solve now for $f \left(9\right)$ using the given function $h \left(t\right) = {t}^{\frac{3}{4}} - 9 \cdot {t}^{\frac{1}{4}}$

$h \left(9\right) = {9}^{\frac{3}{4}} - 9 \cdot {9}^{\frac{1}{4}}$
$h \left(9\right) = {9}^{\frac{3}{4}} - {9}^{\frac{5}{4}}$
$h \left(9\right) = {9}^{\frac{3}{4}} \left(1 - {9}^{\frac{1}{2}}\right)$
$h \left(9\right) = {9}^{\frac{3}{4}} \left(1 - 3\right)$
$h \left(9\right) = {9}^{\frac{3}{4}} \left(- 2\right)$
$h \left(9\right) = - 2 \cdot {9}^{\frac{3}{4}}$

The point $\left(9 , - 2 \cdot {9}^{\frac{3}{4}}\right)$ is a minimum

graph{y=x^(3/4)-9*x^(1/4) [-40, 40, -20, 10]}

God bless....I hope the explanation is useful.