MHB True/False differentiation problem

[Saitama]

Problem:
Let $g(x)$ be twice differentiable function satisfying $g(0)=0$, $g(1)=1$. Then, which of the following is/are correct?

A) there exist distinct $C_1,C_2\in (0,1)$ such that $g'(C_1)+g'(C_2)=2$.

B) there will be atleast one $C$ such that $g'(C)=1$ for $C\in (0,1)$

C) there will be atleast one $C\in (0,1)$ such that $g(C)=C$ for $C\in (0,1)$

D) if $g(\alpha)=\alpha$ for $\alpha \in (0,1)$ then there will exist atleast one $C\in (0,1)$ such that $g''(C)=0$.

Attempt:
The given answer is A, B and D. I have shown that B and D are true and C is false but I am not sure how to begin with proving A. I really don't know where to start.

Any help is appreciated. Thanks!

 

[Sudharaka]

Problem:
Let $g(x)$ be twice differentiable function satisfying $g(0)=0$, $g(1)=1$. Then, which of the following is/are correct?

A) there exist distinct $C_1,C_2\in (0,1)$ such that $g'(C_1)+g'(C_2)=2$.

B) there will be atleast one $C$ such that $g'(C)=1$ for $C\in (0,1)$

C) there will be atleast one $C\in (0,1)$ such that $g(C)=C$ for $C\in (0,1)$

D) if $g(\alpha)=\alpha$ for $\alpha \in (0,1)$ then there will exist atleast one $C\in (0,1)$ such that $g''(C)=0$.

Attempt:
The given answer is A, B and D. I have shown that B and D are true and C is false but I am not sure how to begin with proving A. I really don't know where to start.

Any help is appreciated. Thanks!

Hi Pranav,

Let me give some ideas for proving (A). Basically we have to apply the Mean Value Theorem twice. There exist $C_1\in(0,\,1)$ such that,

\[f'(C_1)=\frac{f(1)-f(0)}{1-0}=1\]

Now again apply the mean value theorem considering the points $C_1$ and $0$. Then you'll get a point $C_2\in (0,\, C_1)$ such that, $f'(C_2)=1$. Hope you can continue. :)

 

[Saitama]

Hi Sudharka! :)

Now again apply the mean value theorem considering the points $C_1$ and $0$. Then you'll get a point $C_2\in (0,\, C_1)$ such that, $f'(C_2)=1$. Hope you can continue. :)

I am not sure but intuitively, I think this isn't necessarily true. For example, consider the graph shown in attachment. There is no $C_2$ in $(0,C_1)$ where $g'(C_2)=1$. What's wrong with my thinking?

 

[Sudharaka]

Hi Sudharka! :)I am not sure but intuitively, I think this isn't necessarily true. For example, consider the graph shown in attachment. There is no $C_2$ in $(0,C_1)$ where $g'(C_2)=1$. What's wrong with my thinking?

Oh sorry. As you see there's a mistake in my previous post. Using the mean value theorem (for the first derivative) on the points $(0,\, C_1)$, we can only get $f''(C_1)=\frac{f'(C_1)-f'(0)}{C_1}$ which doesn't mean a thing. I confused it somehow and thought $f'(C_1)=1$.

 

[Saitama]

Oh sorry. As you see there's a mistake in my previous post. Using the mean value theorem (for the first derivative) on the points $(0,\, C_1)$, we can only get $f''(C_1)=\frac{f'(C_1)-f'(0)}{C_1}$ which doesn't mean a thing. I confused it somehow and thought $f'(C_1)=1$.

I am still not sure where to start. Why are you dealing with the second derivative? Can you please explain a little more?

 

[Evgeny.Makarov]

What happens if you apply the MVT to $[0,1/2]$ and $[1/2,1]$?

 

[ThePerfectHacker]

Define $f(x) = g(x) - g(1-x)$.

 

[Evgeny.Makarov]

Define $f(x) = g(x) - g(1-x)$.

On the one hand, this establishes a stronger fact because not only $g'(C_1)+g'(C_2)=2$, but also $C_1+C_2=1$. On the other hand, it may happen that $C_1=C_2=1/2$, while the problem requires distinct points.

The solutions above require only that $g$ satisfies the MVT on $[0,1]$, i.e., that $f$ is continuous on $[0,1]$ and $g$ is differentiable on $(0,1)$. It is not required that $g$ is twice differentiable. However, a quick-and-dirty way to see that A) is true (say, on a test) is as follows. We know that $g'(c)=1$ for some $c\in(0,1)$ by applying the MVT to $g$. If $g'$ continuous on $[0,1]$ (which happens when $g$ is twice differentiable on $[0,1]$), then it is possible to disturb $g'(c)$ a little so that it becomes $1+\delta$ and $1-\delta$ in two different points. More precisely, $g'(x)$ attains its maximum $M$ and minimum $m$ and we have $m\le1\le M$. Then either $m=M=1$ or $m<1<M$. (If, say, $m=1<M$, then $g$ grows at least as fast as $y=x$, but on some interval, due to continuity of $g'$, it grows strictly faster that $y=x$, so $g(1)>1$.) If $m=M=1$, then $C_1,C_2$ can be any points in $[0,1]$. Otherwise, it is possible to select a $0<\delta<\min(1-m,M-1)$, so $1-\delta$ and $1+\delta$ are in $[m,M]$. By the IVT applied to $g'$ there exist $C_1,C_2$ where $g'(C_1)=1-\delta$ and $g'(C_2)=1+\delta$.

 

[Saitama]

Thanks Evgeny.Makarov and ThePerfectHacker! :)