The critical numbers of the function:
f(t)=7t^(2/3)+t^(5/3)
f'(t)=14/3t^(-1/3)+(5/3)t^(2/3)
I got 0 as a critical number but it is hard for me to factor the f'(t). Whats the other critical number? Thank you.
If 14/3t^(-1/3)+(5/3)t^(2/3)= 0, then
t^(-1/3) [14/3 + (5/3)t] = 0
t = 0 or -14/5
To find the critical numbers of a function, we need to find the values of t where the derivative is equal to zero or undefined. In your case, you correctly found that f'(t) = 14/3t^(-1/3) + (5/3)t^(2/3).
To find the critical numbers, we set f'(t) equal to zero and solve for t:
14/3t^(-1/3) + (5/3)t^(2/3) = 0
Now, let's combine the terms with different powers of t:
14t^(2/3) = -5t^(-1/3)
Now, we can eliminate the fractions by multiplying both sides by 3t^(1/3):
42t = -15
Simplifying further, we have:
t = -15 / 42 = -5 / 14
So, one critical number of the function f(t) is t = -5 / 14.
To find the other critical number, we need to consider when the derivative is undefined. In this case, the derivative is undefined if t^(-1/3) = 0, which means t = 0.
Therefore, the other critical number of the function f(t) is t = 0.
To summarize, the critical numbers of the function f(t) = 7t^(2/3) + t^(5/3) are t = -5/14 and t = 0.