1. ## Mean value theorem

Hello Forumites,
1) Suppose that $-1\leq f'x(x) \leq 3$ for all x. Author wants me to find similar lower and upper bounds for the expression f(5)-f(3).

2)Suppose g(x) is a function that is differentiable for all x. Let h(x0) be a new function defined by h(x)=g(x)+g(2-x). Author wants me to prove that h'(x)has a root in a interval (0,2)
Solutions
I didn't understand both of these questions. Any forumites may reply the answers to both these questions.

2. ## Re: Mean value theorem

please clarify what this is saying

$\displaystyle -1\leq f'x(x) \leq 3$

3. ## Re: Mean value theorem

Originally Posted by Idea
please clarify what this is saying

$\displaystyle -1\leq f'x(x) \leq 3$
Hello,
That's what I didn't understand. Whatever I saw in the PDF file I posted it here.

4. ## Re: Mean value theorem

For 1, I am assuming the PDF has a typo. It should read:

$$-1\le f'(x) \le 3$$

WLOG, assume $f(3)=0$. To get the lower bound of the expression, assume the derivative is always -1 (since derivative is constant, this is a line, so $f(5)$ is to the right two of $f(3)$, and it must also be down two. To get the upper bound, assume it is always 3. This gives you:

$$-2\le f(5)-f(3) \le 6$$

For 2: I will assume another typo. Take the derivative of $h(x)$. Plug in $x=1$. What do you get?

5. ## Re: Mean value theorem

Originally Posted by SlipEternal
For 1, I am assuming the PDF has a typo. It should read:

$$-1\le f'(x) \le 3$$

WLOG, assume $f(3)=0$. To get the lower bound of the expression, assume the derivative is always -1 (since derivative is constant, this is a line, so $f(5)$ is to the right two of $f(3)$, and it must also be down two. To get the upper bound, assume it is always 3. This gives you:

$$-2\le f(5)-f(3) \le 6$$

For 2: I will assume another typo. Take the derivative of $h(x)$. Plug in $x=1$. What do you get?
Hello,
1)Note that all conditions of the Mean Value Theorem are satisfied. To get the bounds use the fact, that for some $c \in (1,3),$ f(5)-f(3)=2f'(c).
2)Note that h(2)-h(0)=0 and apply the Mean Value Theorem for the function h on the closed interval [0,2].

6. ## Re: Mean value theorem

Originally Posted by SlipEternal
For 1, I am assuming the PDF has a typo. It should read:

$$-1\le f'(x) \le 3$$

WLOG, assume $f(3)=0$. To get the lower bound of the expression, assume the derivative is always -1 (since derivative is constant, this is a line, so $f(5)$ is to the right two of $f(3)$, and it must also be down two. To get the upper bound, assume it is always 3. This gives you:

$$-2\le f(5)-f(3) \le 6$$

For 2: I will assume another typo. Take the derivative of $h(x)$. Plug in $x=1$. What do you get?
Hello,

7. ## Re: Mean value theorem

If $h(x) = g(x)+g(2-x)$ then $h'(x) = g'(x)+g'(2-x)\dfrac{d}{dx}\left(2-x\right) = g'(x) -g'(2-x)$. Plugging in $x=1$ gives $h'(1) = g'(1)-g'(2-1) = g'(1)-g'(1) = 0$.

If you have not yet learned the chain rule, then this result is not meaningful to you. In that case, the book is using the Mean Value Theorem instead of the Chain Rule.

$$h(2)-h(0) = g(2)+g(2-2)-(g(0)+g(2-0)) = g(2)+g(0)-g(0)-g(2) = 0$$

Also, by the Mean Value Theorem:

$$h(2)-h(0) = ch'(c)$$

for some value of $c \in (0,2)$. Therefore, there exists $c \in (0,2)$ such that $ch'(c) = 0$. Either $c=0$ or $h'(c)=0$. But, $c \in (0,2)$ means that $c\neq 0$. Therefore, $h'(c) = 0$.

8. ## Re: Mean value theorem

Originally Posted by SlipEternal
If $h(x) = g(x)+g(2-x)$ then $h'(x) = g'(x)+g'(2-x)\dfrac{d}{dx}\left(2-x\right) = g'(x) -g'(2-x)$. Plugging in $x=1$ gives $h'(1) = g'(1)-g'(2-1) = g'(1)-g'(1) = 0$.

If you have not yet learned the chain rule, then this result is not meaningful to you. In that case, the book is using the Mean Value Theorem instead of the Chain Rule.

$$h(2)-h(0) = g(2)+g(2-2)-(g(0)+g(2-0)) = g(2)+g(0)-g(0)-g(2) = 0$$

Also, by the Mean Value Theorem:

$$h(2)-h(0) = ch'(c)$$

for some value of $c \in (0,2)$. Therefore, there exists $c \in (0,2)$ such that $ch'(c) = 0$. Either $c=0$ or $h'(c)=0$. But, $c \in (0,2)$ means that $c\neq 0$. Therefore, $h'(c) = 0$.
Hello,
But h'(x) can be written as g'(x)+g'(2)-g'(x)=0 without chain rule.

9. ## Re: Mean value theorem

Originally Posted by Vinod
Hello,
But h'(x) can be written as g'(x)+g'(2)-g'(x)=0 without chain rule.
No, it cannot be.

Example:

$$g(x) = x^2$$

$$h(x) = g(x)+g(2-x) = x^2+(2-x)^2 = 2x^2-4x+4$$

$$h'(x) = 4x-4$$

$$g'(x)+g'(2)-g'(x)=g'(2) = 4 \neq h'(x)$$