Find the critical numbers of the function.

h(t)=t^3/4-6^1/4

I think there's a typo there. Anyway, just find where h'=0.

oh, sorry about that!

h(t)=t^3/4-6t^1/4

well, dh/dt = 3t^2(11∜t-8) / 8(2-3∜t)^2

We want dh/dt=0, so t=0 or t=(8/11)^4

h(t) and dh/dt are undefined where t=(2/3)^4

To find the critical numbers of a function, we need to find the values of t where the derivative of the function, h'(t), equals zero or is undefined.

Let's find the derivative of the function h(t) first.

h(t) = t^(3/4) - 6^(1/4)

To differentiate the function, we'll use the power rule and chain rule:

h'(t) = (3/4)t^(3/4 - 1) - (1/4)(6^(1/4 - 1))(6)^(1/4 - 1)

Simplifying this expression:

h'(t) = (3/4)t^(-1/4) - (1/4)(6^(-3/4))(6^(-3/4))

h'(t) = (3/4)t^(-1/4) - (1/4)(6^(-3/4))(1/6^3/4)

h'(t) = (3/4)t^(-1/4) - (1/4)(1/6^(3/4))

Now, we'll set h'(t) equal to zero and solve for t:

(3/4)t^(-1/4) - (1/4)(1/6^(3/4)) = 0

(3/4)t^(-1/4) = (1/4)(1/6^(3/4))

t^(-1/4) = (1/4)(1/6^(3/4))/(3/4)

t^(-1/4) = (1/4)(1/6^(3/4))*(4/3)

t^(-1/4) = (1/6^(3/4))*(1/3)

Now, we'll take the reciprocal of both sides to solve for t:

t^(1/4) = (3/6^(3/4))

t = (3/6^(3/4))^(4/1)

Simplifying this expression further:

t = (3^4)/(6^3)

t = 81/216

Therefore, the critical number of the function h(t) is t = 81/216.