Question

step 1
for ( h ( t ) = t ^ { 3 / 4 } - 8 t ^ { 1 / 4 } ), we have
step 2
( h ^ { prime } ( t ) = \frac { 3 } { 4 } t ^ { - \frac { 1 } { 4 } } - 2 t ^ { - \frac { 3 } { 4 } } )
( = \frac { 3 sqrt { t } - 8 } { 4 sqrt 4 { t ^ { 3 } } } )
step 2
critical numbers occur where ( h ^ { prime } ( t ) = 0 ) and where ( h ^ { prime } ( t ) ) is undefined.
( 0 = h ^ { prime } ( t ) = \frac { 1 } { 4 } t ^ { - 3 / 4 } ( 3 t ^ { 1 / 2 } - 8 ) = \frac { 3 t ^ { 1 / 2 } - 8 } { 4 sqrt 4 { t ^ { 3 } } } )
when the numerator is 0, which occurs at the following values. (enter your answers as a comma-separated list.)
( t = \frac { 64 } { 9 } )
( t = \frac { 64 } { 9 } )
step 3
( h ^ { prime } ( t ) = \frac { 3 t ^ { 1 / 2 } - 8 } { 4 sqrt 4 { t ^ { 3 } } } ) is undefined when the denominator is 0, which occurs where ( t ) is equal to the following values. (enter your answers as a comma-separated list.)
( t = )

Explanation:

Step1: Find the derivative of \(h(t)\)

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(h(t)=t^{3/4}-8t^{1/4}\), we have \(h^\prime(t)=\frac{3}{4}t^{\frac{3}{4}-1}-8\times\frac{1}{4}t^{\frac{1}{4}-1}=\frac{3}{4}t^{-\frac{1}{4}} - 2t^{-\frac{3}{4}}=\frac{3t^{\frac{1}{2}}-8}{4t^{\frac{3}{4}}}\)

Step2: Find the values where \(h^\prime(t) = 0\)

Set the numerator equal to \(0\): \(3t^{\frac{1}{2}}-8 = 0\). Then \(3\sqrt{t}=8\), so \(\sqrt{t}=\frac{8}{3}\), and \(t=\frac{64}{9}\)

Step3: Find the values where \(h^\prime(t)\) is undefined

Set the denominator equal to \(0\): \(4t^{\frac{3}{4}} = 0\), which gives \(t = 0\)

Answer:

\(0,\frac{64}{9}\)