Using the first derivative test, identify and label the extrema of the given function:

Question

g(t)= t√(17-t) , t < 17

Answer 1

First, get the derivative of g(t) using the product rule.

g(t)= t(17-t)^1/2 , t < 17

g'(t) = (17-t)^1/2 + t(1/2)(17-t)^-1/2 (-1)

g'(t) = (17-t)^1/2 - t /(2(17-t)^1/2)

g'(t) = (2(17-t) - t) /(2(17-t)^1/2)

g'(t) = (34-2t - t) /(2(17-t)^1/2)

g'(t) = (34 - 3t) /(2(17-t)^1/2)

Then solve for t when g'(t) = 0.

0 = (34 - 3t) /(2(17-t)^1/2)

0 = 34 - 3t

3t = 34

t = 34/3 ≈ 11.33

plugin t = 34/3 in g(t):

g(34/3) = (34/3)(17 - 34/3)^1/2

g(34/3) = 34√51/ 9 ≈ 26.98

Therefore the extrema (which is only one maximum and no minimum) is at point

(34/3 , 34√51/ 9)