MHB How to prove such value for a derivative?

[FallArk]

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 $$\in$$ (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

 

[HallsofIvy]

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.

 

[FallArk]

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.

One more question, we did not learn that property in class. Is it possible to prove that f' is a continuous function? so I can use the intermediate value theorem to say that f'(c) = 1/2 exists

 

[HallsofIvy]

No, because for general f, f' is not necessarily continuous! (I said that in my original post). An example is f(x)= \frac{sin(x)}{x^2} if x is not 0, 0 if x is 0.
For x not 0, the derivative is f'(x)= \frac{x^2 cos(x)- 2x sin(x)}{x^4}= \frac{x cos(x)- 2 sin(x)}{x^3} 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.

 

[Opalg]

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 $$\in$$ (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

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]

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 $f'(c) = 0.5.$

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...

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No, because for general f, f' is not necessarily continuous! (I said that in my original post). An example is f(x)= \frac{sin(x)}{x^2} if x is not 0, 0 if x is 0.
For x not 0, the derivative is f'(x)= \frac{x^2 cos(x)- 2x sin(x)}{x^4}= \frac{x cos(x)- 2 sin(x)}{x^3}

if x is not 0, and 0 at 0. Thus f&#039;(x) exists for all x but since the limit of f&#039;(x) as x goes to 0 is not 0, f&#039;(x) is not continuous at x= 0.[/QUOTE]<br /> <br /> I know it does not need to continuous, so is it okay to simply state that still by MVT, f&#039;(c)= 1/2 exists? I&#039;m asking because our textbook did not mention any of that

 

[Opalg]

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...

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]

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 $[d,3].$

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]

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

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]

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 $g'(c) = 0.$

Duh... Of course! How did I forget! Thank you!