by taking the first derivative, where is it equal to zero?
y=x^2-2-cosx on interval [-1,3]
is one of them zero, but since it isnt in the interval, there arent any critical numbers?
2006-11-08 09:17:40 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
ohh haha yea..my b it is in the interval..=]
thanxxxx!!!
2006-11-08 09:37:37 · update #1
Wait a minute, did you want to know when the function is zero or when its derivative is zero?
For derivative:
y' = 2x + sin x, right?
It looks like there's one place where it's 0. Requires and numerical solution.
Well, it's 0 at x=0, right?
I assume that x is in radians. 1 radian is about 57 deg. As x decreases from 0 both parts become more negative.
It looks like x = 0 is the only zero of the dirivative.
2006-11-08 09:28:57 · answer #1 · answered by modulo_function 7 · 3⤊ 2⤋
The derivative is
y' = 2x + sinx
2x + sinx = 0 when x = 0
0 is in the interval [-1,3]
2006-11-08 09:29:43 · answer #2 · answered by MsMath 7 · 3⤊ 3⤋
-the derivative is y' = 2x + sinx
2x + sinx = 0, when x = 0
2006-11-09 07:25:48 · answer #3 · answered by Anonymous · 1⤊ 1⤋