HallsofIvy said:There are two things you need, both of which depend on the "mean value theorem". First, directly from the mean value theorem, since the slope of the straight line from (1, 1) to (2, 3) is 1, there exist a value, a, between 1 and 2, such that f'(a)= 1 and, since the slope of the straight line from (2, 3) to (3, 3) is 0, there exist a point, b, between 2 and 3, such that f'(b)= 0.
Now, one can prove, using the mean value theorem, that, although f' is not necessarily continuous, it does have the "intermediate value property" (so f' is a "Darboux function": http://www.math.wvu.edu/~kcies/teach/Fall03Spr04/451NadlerText/451Nadler101-120.pdf) that between a and b, f' takes on all value between f'(a) and f'(b). Here, f' takes on all values between 0 and 1 so there exist c, between a and b and so between 0 and 3, such that f'(c)= 1/2.
Let $g(x) = f(x) - 0.5x$. Can you use the given information (together with the intermediate value theorem and Rolle's theorem) to show that $g'(c) = 0$ for some point $c\in (1,3)$? It will then follow that $f'(c) = 0.5.$FallArk said:Prove that if f is a differentiable function on R such that f(1) = 1, f(2) = 3, f(3) = 3. There is a c \(\displaystyle \in\) (1 , 3) such that f'(c) = 0.5
I think the mean value theorem should be used, but I can't figure out how to prove such value exists
Opalg said:Let $g(x) = f(x) - 0.5x$. Can you use the given information (together with the intermediate value theorem and Rolle's theorem) to show that $g'(c) = 0$ for some point $c\in (1,3)$? It will then follow that $f'(c) = 0.5.$
if x is not 0, and 0 at 0. Thus f'(x) exists for all x but since the limit of f'(x) as x goes to 0 is not 0, f'(x) is not continuous at x= 0.[/QUOTE]HallsofIvy said:No, because for general f, f' is not necessarily continuous! (I said that in my original post). An example is [tex]f(x)= \frac{sin(x)}{x^2}[/tex] if x is not 0, 0 if x is 0.
For x not 0, the derivative is [tex]f'(x)= \frac{x^2 cos(x)- 2x sin(x)}{x^4}= \frac{x cos(x)- 2 sin(x)}{x^3}
That's where the intermediate value theorem comes in. Use it to show that there is a point $d$ between $1$ and $2$ at which $g(d) = g(3)$. Then you can use Rolle's theorem on the interval $[d,3].$FallArk said:I don't think Rolle's theorem would work though, then g(1) would need to be equal to g(3), but they are not...
Opalg said:That's where the intermediate value theorem comes in. Use it to show that there is a point $d$ between $1$ and $2$ at which $g(d) = g(3)$. Then you can use Rolle's theorem on the interval $[d,3].$
Since $g(x) = f(x) - \frac12x$, it follows that $g(1) = \frac12$, $g(2) = 2$ and $g(3) = \frac32$. But $\frac12<\frac32<2$, so by the intermediate value theorem for $g(x)$ on the interval $[1,2]$, there exists $d\in [1,2]$ such that $g(d) = \frac32$. Thus $g(d) = g(3) = \frac32$ and now you can apply Rolle's theorem to $g(x)$ on the interval $[d,3]$ to deduce that there exists $c\in[d,3]$ such that $g'(c) = 0.$FallArk said:I don't think it is possible, because IVT requires that for some d (in this case 3) between g(a) and g(b), there is a point c where g(c) = d. but the function g(x) gives the value of 1/2 and 2 on [1,2]
Did I miss something? Maybe it could work and I forgot something important
Opalg said:Since $g(x) = f(x) - \frac12x$, it follows that $g(1) = \frac12$, $g(2) = 2$ and $g(3) = \frac32$. But $\frac12<\frac32<2$, so by the intermediate value theorem for $g(x)$ on the interval $[1,2]$, there exists $d\in [1,2]$ such that $g(d) = \frac32$. Thus $g(d) = g(3) = \frac32$ and now you can apply Rolle's theorem to $g(x)$ on the interval $[d,3]$ to deduce that there exists $c\in[d,3]$ such that $g'(c) = 0.$
A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It measures how much a function changes as its input changes.
Proving the value of a derivative is important because it allows us to understand the behavior of a function and make accurate predictions about its future values. It also helps us to solve complex mathematical problems and make informed decisions in various fields such as economics, physics, and engineering.
The most common method to prove the value of a derivative is through the use of the limit definition of a derivative. This involves taking the limit of a function as the change in its input approaches zero. Other methods include using differentiation rules, such as the power rule or chain rule, and using graphical representations of a function.
A function and its derivative are closely related, as the derivative represents the instantaneous rate of change of the function. In other words, the derivative tells us how much the output of a function changes for a small change in its input. It is also important to note that the derivative of a function is also a function itself.
Not all functions have a well-defined derivative. Some functions, such as discontinuous or non-differentiable functions, do not have a derivative at certain points. In these cases, it is not possible to prove the value of the derivative. However, for most continuous and differentiable functions, it is possible to prove the value of the derivative using various mathematical techniques.