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(11βt-8) / 8(2-3βt)^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.