h' = 3/4 t^-1/4 - 3/4 t^-3/4
= 3/4 t^-3/4 (t^1/2 - 1)
so, when is h' = 0?
h(t) = t^(3/4) − 3t^(1/4)
= 3/4 t^-3/4 (t^1/2 - 1)
so, when is h' = 0?
Let's find the derivative of h(t) first.
h(t) = t^(3/4) - 3t^(1/4)
To find the derivative, we'll use the power rule of differentiation. The power rule states that if we have a term of the form t^n, then the derivative of t^n is n * t^(n-1).
Using the power rule, we can find the derivative of each term in h(t):
h'(t) = (3/4)t^(-1/4) - (3/4)t^(-3/4)
To find the critical numbers, we need to set the derivative equal to zero and solve for t. Let's set h'(t) = 0:
(3/4)t^(-1/4) - (3/4)t^(-3/4) = 0
Multiplying both sides by 4/3 to clear the fractions, we get:
t^(-1/4) - t^(-3/4) = 0
Adding t^(-3/4) to both sides, we get:
t^(-1/4) = t^(-3/4)
Since the bases are the same, we can equate the exponents:
-1/4 = -3/4
This equation has no solution.
So, there are no critical numbers for the function h(t) = t^(3/4) - 3t^(1/4). Therefore, the answer is DNE (Does Not Exist).
First, let's find the derivative of h(t):
h'(t) = (3/4)t^(-1/4) - (3/4)t^(-3/4)
Next, let's set h'(t) equal to zero and solve for t:
(3/4)t^(-1/4) - (3/4)t^(-3/4) = 0
Multiplying through by 4 to clear the denominators, we get:
3t^(-1/4) - 3t^(-3/4) = 0
Dividing through by 3, we have:
t^(-1/4) - t^(-3/4) = 0
Next, let's factor out t^(-3/4) from both terms:
t^(-3/4)(t^(1/2) - 1) = 0
Setting each factor equal to zero, we have:
t^(-3/4) = 0 and t^(1/2) - 1 = 0
Since t^(-3/4) = 0 does not have a real solution (it is not defined when t = 0), we can ignore it.
Solving t^(1/2) - 1 = 0, we get:
t^(1/2) = 1
Squaring both sides, we have:
t = 1
Therefore, the only critical number of the function h(t) = t^(3/4) - 3t^(1/4) is t = 1.
So, the critical number is 1.