What are the critical numbers of f(t)=3t^(2/3)+t^(5/3)
To find the critical numbers of the function f(t) = 3t^(2/3) + t^(5/3), we need to follow these steps:
Step 1: Find the derivative of f(t).
Start by finding the derivative of f(t) with respect to t. The derivative will help us determine where the function's slope is zero or undefined, which are the critical points.
The derivative of f(t) is found using the power rule and chain rule:
f'(t) = (2/3) * 3t^((2/3) - 1) + (5/3) * t^((5/3) - 1)
Simplifying it further:
f'(t) = 2t^(-1/3) + (5/3)t^(2/3 - 1)
Step 2: Determine where the derivative is equal to zero or undefined.
Set the derivative equal to zero and solve for t to find the values where the slope is zero.
2t^(-1/3) + (5/3)t^(2/3 - 1) = 0
Simplifying further, we get:
2t^(-1/3) + (5/3)t^(-1/3) = 0
To solve this equation, we can factor out the common term t^(-1/3):
t^(-1/3)(2 + (5/3)) = 0
Simplifying again, we find:
t^(-1/3)(16/3) = 0
Since t^(-1/3) cannot be zero, we set the second part of the equation to zero:
16/3 = 0
However, there is no real value of t that can make 16/3 equal to zero, so there are no critical points when the derivative is zero.
Step 3: Check for points of discontinuity.
To identify any points of discontinuity or where the derivative is undefined, let's analyze the original function.
f(t) = 3t^(2/3) + t^(5/3)
The original function is defined for all real values of t, so there are no points of discontinuity.
Therefore, there are no critical numbers for the function f(t) = 3t^(2/3) + t^(5/3).
Those would be the values of t for which the derivative f'(t) = 0 is zero,
The derivative is
f'(t) = 2 t^(-1/3) + (5/3) t^(2/3)
That is zero when
(5/6)t^(2/3) = -t(-1/3)
5/6 t = -1
t = -6/5