9) Find all critical numbers of the function
f(t)=9t^2/3+t^5/3
To find the critical numbers of a function, we need to start by finding its derivative and then solving for when the derivative is equal to zero or undefined.
Let's find the derivative of the function f(t) = 9t^(2/3) + t^(5/3):
To find the derivative of the function, we will need to use the power rule and chain rule.
Step 1: Find the derivative of the first term, 9t^(2/3):
The derivative of t^(2/3) with respect to t is (2/3) * t^((2/3) - 1) = (2/3) * t^(-1/3).
Multiplying by the coefficient 9, the derivative of 9t^(2/3) is 9 * (2/3) * t^(-1/3) = 6t^(-1/3).
Step 2: Find the derivative of the second term, t^(5/3):
The derivative of t^(5/3) with respect to t is (5/3) * t^((5/3) - 1) = (5/3) * t^(2/3).
Therefore, the derivative of t^(5/3) is (5/3) * t^(2/3).
Step 3: Combine the derivatives of both terms:
The derivative of f(t) = 9t^(2/3) + t^(5/3) is the sum of the derivative of each term, which gives us the overall derivative:
f'(t) = 6t^(-1/3) + (5/3) * t^(2/3).
Now, let's solve for when the derivative is equal to zero or undefined. In other words, we need to find the critical points.
Set the derivative equal to zero and solve for t:
6t^(-1/3) + (5/3) * t^(2/3) = 0.
We can multiply through by 3 to eliminate the fraction:
18t^(-1/3) + 5 * t^(2/3) = 0.
Next, we need to rearrange the equation to make it easier to solve:
5 * t^(2/3) = -18t^(-1/3).
Divide both sides by t^(2/3):
5 = -18/t.
Multiply both sides by t:
5t = -18.
Divide both sides by 5:
t = -18/5.
So the critical number of the function f(t) = 9t^(2/3) + t^(5/3) is t = -18/5.
f(t) = 1/3 (9t^2 + t^5)
f'(t) = 1/3 (18t + 5t^4)
= 1/3 t(18+5t^4)
f'=0 where x=0 or ±∜3.6