CompuChip said:There is no product rule to be applied.
sin(2x) is something different than sin(x) * 2x.
If you call f(x) = sin(x), this is easier to see (f(2x) and 2 x f(x) are completely different things, and the derivative of f(2x) is not 2f(x) + 2x f'(x), or something like that).
What you are looking for is the chain rule.
Mark44 said:What does Rolle's Theorem say? You have a function, f(x) = sin(2x) for which f(0) = f(pi/2) = 0, and f is a differentiable function (you found its derivative).
What does Rolle's Theorem conclude?Jimmy84 said:So the zeros are 0 and pi/2 . The function is continuous on [0,pi/2] and differentiable on
(0,pi/2)
Why would you want to evaluate f'(0)?Jimmy84 said:Well I graphed the function and the maximum extreme is obiously 1 but if I evaluate
2cos(2x) that is the derivtive of f by 0 I get
2cos(0) = 2
what am I doing wrong?
ideasrule said:You've shown that the derivative of sin(2x) is 2 at x=0. Nothing in math says that the derivative can't be larger than the maximum value of the function.
Draw out a rough sketch of sin(2x). What's the slope of the tangent line equal to at the highest and lowest points?
Note that you should be able to solve this question intuitively, without using calculus. What's the maximum and minimum values of any sine function?
This is not necessarily true. For example, the minimum value of y = f(x) = |x| is 0, and it occurs at x = 0, but f'(x) is never equal to 0 for this function.Jimmy84 said:The derivative of any maximum or minimum equals to 0
Close. Rolle's Thm says that there is a number c in the open interval (0, pi/2) such that f'(c) = 0.Jimmy84 said:and the pourpose of Rolle's theorem as far as I know is to show that there is a number c in the interval [0,pi/2] such that f '(c) = 0
Jimmy84 said:Im being asked to find c as well. how can I find c ?
I think c is called a critical point as well.
Mark44 said:This is not necessarily true. For example, the minimum value of y = f(x) = |x| is 0, and it occurs at x = 0, but f'(x) is never equal to 0 for this function.
Close. Rolle's Thm says that there is a number c in the open interval (0, pi/2) such that f'(c) = 0.
That means that for some number 0 < c < pi/2, 2cos(2c) = 0. Can you solve this equation?
No, f(x) = |x| is defined for all values of x, but for this function, f' is not defined at x = 0.Jimmy84 said:Thanks, that's right I think that f(x) = |x| is undefined at x= 0 .
Jimmy84 said:no, I can't solve that equation this is kind of new to me. that's the main issue I am having with this problem so I don't have a clue. I don't remember ever solving variable equations like that.
I guess that by guessing it could be factorized like this
2 (cos (c)) = 0
but that's about it.
Yes - it's not right.Jimmy84 said:I don't think that this is right but anyway
2cos(2c) = 0
2c = 1/2 cos x = c = 1/4 cos x
but I don't understand where did pi came from in c = 1/4pi which is the original result of the book.
Can you give me a hint?
Mark44 said:Yes - it's not right.
2cos(2c) = 0
==> cos(2c) = 0
Now, I'm hopeful that you have seen the graph of y = cos(x).
For what values of x is cos(x) = 0? One of them is pi/2 and there are lots and lots of values.
The same values of x for which cos(x) = 0 are the same values of 2c, so what are the values of c?
Rolle's Theorem is a mathematical theorem that states that if a function is continuous on a closed interval and differentiable on the open interval, and the function's values at the endpoints of the interval are equal, then there exists at least one point within the interval where the derivative of the function is equal to zero.
Rolle's Theorem can be used to find extremes of a function by first finding the critical points of the function, which are the points where the derivative is equal to zero. These critical points correspond to the maximum and minimum values of the function within the given interval.
To apply Rolle's Theorem to the function f(x) = sin 2x, you need to first determine the interval over which you want to find the extremes. Then, take the derivative of the function, which in this case is f'(x) = 2cos 2x. Set the derivative equal to zero and solve for x to find the critical points. Finally, plug these critical points back into the original function to find the corresponding y-values at the extremes.
No, Rolle's Theorem can only be used to find the extremes of a function within a given interval. It does not guarantee that all the extremes of a function will be found, as there may be critical points outside of the given interval or the function may have other types of extremes such as inflection points.
Rolle's Theorem can only be applied to continuous and differentiable functions, so it may not be applicable to all types of functions. Additionally, it can only find the extremes within a given interval, so it may not provide a complete picture of the function's behavior. It is also important to note that finding the critical points does not guarantee that they are actual extremes of the function, as they could be points of inflection or points where the derivative is undefined.