Use the Mean Value Theorem to show:
-x < sinx < x for all x > 0
(and where I have the inequalities, it's supposed to be less than or equal to for the first two and greater than or equal to for the last one, I just didn't know how to type it)
Well, well, well! Looks like you want to see some mathematical clownery with the Mean Value Theorem. Brace yourself for some fun!
The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number "c" in the interval (a, b) such that the derivative of the function at that point is equal to the average rate of change of the function over the interval.
Now, let's apply this theorem to the sine function, shall we? We're interested in the interval (0, x). The sine function is certainly continuous on this interval, and it's also differentiable on the open interval (0, x) for any positive value of x. So, we're good to go!
Let's consider the function f(x) = sin(x) on the interval [0, x]. By the Mean Value Theorem, we know that there exists a number "c" in the interval (0, x) where:
f'(c) = (f(x) - f(0)) / (x - 0)
Since f(0) = 0, we can simplify it as:
f'(c) = f(x) / x
Let's analyze the inequality for the derivative f'(c):
-x < sin(x) < x
Now, let's divide all parts of this inequality by x:
-1 < sin(x) / x < 1
And there you have it! Using the Mean Value Theorem and a sprinkle of mathematical shenanigans, we've shown that:
-x < sin(x) < x for all x > 0
Remember, laughter and math go hand in hand, so keep smiling!