That is, 0
x - 2 is not a inequality and so the question does not make any sense.
f(x) is decreasing on the interval on which f'(x) is negative. So we want: (x2-2)/x<0 For this to be true either the numerator or the denominator (but not both) must be negative. On the interval x>0, the numerator is negative for 0<x<sqrt(2) and the denominator is positive for all x>0. On the interval x<0, the denominator is negative for all values on this interval. The numerator is positive on this interval for x<-sqrt(2). So, f' is negative (and f is decreasing) on the intervals: (-infinity, -sqrt(2)), (0, sqrt(2))
You would write it as 93
Let g(x) = interval [0, x] of sin t dt, and f(t) = sin t. Since f(t) is a continuous function, the part one of the Fundamental Theorem of Calculus gives, g'(x) = sin x = f(x) (the original function). If you are interested in the interval [x, 0] of sin t dt, then just put a minus sign in front of the integral and interchange places of 0 and x. So that, g(x) = interval [x, 0] of sin t dt = -{ interval [0, x] of sin t dt}, then g'(x) = - sin x.
0 < x < +INF is the same as: (0, +INF)
(0, infinity) Note that you should use the infinity sign instead of the word (the sideways 8).
A function is positive on an interval, say, the interval from x=a to x=b, if algebraically, all the y-coordinate values are positive on this interval; and graphically, the entire curve or line lies above the x-axis.on this interval.
That is, 0
-3
32
[0, x] assuming both 0 and x are included. If neither are included it would be (0, x). If 0 is included but x is not then it is [0, x). If x is included but 0 is not then it is (0, x].
x - 2 is not a inequality and so the question does not make any sense.
6.29 x 107 is already in scientific notation.
If we write 0.00025006 in scientific notation, we write it like this: 2.5006 x 10-4.
2x2x2x3x3x5x5 in exponential notation = 23 x 32 x 52
Expanded Notation of 14 = (1 x 10) + (4 x 1)