Find the critical numbers of the function. (Enter your answers as a comma-separated list.)

Question

Answer 1

g(t) = tsqr(5-t)

g'(t) = (t times derivative of sqr(5-t)) plus sqr(5-t) = (10-3t)/2sqr(5-t) = 0

10-3t = 0

t = 10/3 = 3 1/3

you can find 3 1/3 as a critical point also by graphing the function. It is a local and absolute maximum.

5 is also a critical point, a local minimum, found by graphing or by finding where the derivative is undefined

Answer 2

Critical points occur where the derivative of g(t) = 0.

The derivative of √(5 – t) = –1/[2√(5 – t)]

1. Use this and the product rule to find the derivative g'(t) = t√(5 – t)
2. Set g'(t) = 0
3. Solve for t

Hope this helps. If you have any questions (or get stuck) just add a comment.

(Hint: the critical point is a number between 3 and 4.)