find the critical numbers of the function 2t^(2/3)+t^(5/3)
Those would be where the derivative is zero.
(4/3)t^(-1/3) + (5/3) t(2/3) = 0
t^(-1/3)*[(4/3) + (5/3)t] = 0
t = 0 or -4/5
To find the critical numbers of the function 2t^(2/3) + t^(5/3), we need to determine where the derivative of the function is equal to zero or does not exist. The critical numbers occur at these points. Let's find the derivative of the function:
f(t) = 2t^(2/3) + t^(5/3)
To find the derivative, we can use the power rule for differentiation:
f'(t) = (2/3)(2)t^(-1/3) + (5/3)(t^2/3)
Simplifying this expression, we get:
f'(t) = (4/3)t^(-1/3) + (5/3)t^(2/3)
Now, set this derivative equal to zero and solve for t:
(4/3)t^(-1/3) + (5/3)t^(2/3) = 0
To solve this equation, we can multiply through by 3t^(1/3) to get rid of the fractions:
4t + 5t^2 = 0
Now, factor out t from the equation:
t(4 + 5t) = 0
Setting each factor equal to zero, we have:
t = 0 and 4 + 5t = 0
From the first equation, t = 0, we have one critical number.
From the second equation, 4 + 5t = 0, we solve for t:
5t = -4
t = -4/5
Therefore, there are two critical numbers for the function 2t^(2/3) + t^(5/3): t = 0 and t = -4/5.