let y = h
dy/dh = 1
let y = √h = h^(1/2)
dy/dh = (1/2)h^(-1/2 = 1/(2√h)
so when is 1 = 2(1/2)/(2√h)
1 = 1/(2√h)
2√h = 1
√h=1/2
h = 1/4
At what value of h is the rate of increase of √h twice the rate of increase of h?
(a) 1/16
(b) 1/4
(c) 1
(d) 2
(e) 4
2 answers
answer is a = 1/16