f(x) = (10)/ (x^2 + 1) on the interval [-1,2].
2006-08-18 00:48:18 · 5 answers · asked by tjhauck2001 2 in Science & Mathematics ➔ Mathematics
Nagpapatawa ba kayo?
f'(x) = -20x / (x^2 + 1)^2 = 0
x = 0
If x = 0,
f(x) = 10
If x = -1,
f(x) = 5
If x = 2,
f(x) = 2
Therefore,
absolute maximum is 10 at x = 0
absolute minimum is 2 at x = 2
2006-08-18 00:55:49 · answer #1 · answered by Joe Mkt 3 · 2⤊ 0⤋
take inflexion points and 0 to check it
put the value of x as -1, 0, and 2 and find it
max is at x = 0 f(x) = 10
min at x = 2 f(x) = 2
2006-08-18 09:38:04 · answer #2 · answered by DG 3 · 0⤊ 1⤋
max f(x) = 10 at x= 0
and min f(x) = 2 at x= 2
2006-08-18 08:24:30 · answer #3 · answered by DeAd MaN 4 · 0⤊ 0⤋
max is when x = 0 and min is when x=2.
2006-08-18 07:58:01 · answer #4 · answered by Goose 2 · 0⤊ 0⤋
2 & 5, respectively.
2006-08-18 07:57:36 · answer #5 · answered by nilay_space 2 · 0⤊ 1⤋