g(x)= x sqrt (9-x^2) on [0,3]
Max:
3 answers
see the "math Calculas" post by Chloe below
thx
sorry - I messed that one up
g(x) = x√(9-x^2)
g' = (9-2x^2)/√(9-x^2)
local max at x = 3/√2
g(0) = 0
g(3) = 0
so abs max is 9/2
abs min = 0
g(x) = x√(9-x^2)
g' = (9-2x^2)/√(9-x^2)
local max at x = 3/√2
g(0) = 0
g(3) = 0
so abs max is 9/2
abs min = 0
1 answer