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Thread: Mean value theorem

  1. #1
    Senior Member Vinod's Avatar
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    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.
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  2. #2
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    Re: Mean value theorem

    please clarify what this is saying

    $\displaystyle -1\leq f'x(x) \leq 3$
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  3. #3
    Senior Member Vinod's Avatar
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    Re: Mean value theorem

    Quote Originally Posted by Idea View Post
    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.
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    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?
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    Senior Member Vinod's Avatar
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    Re: Mean value theorem

    Quote Originally Posted by SlipEternal View Post
    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,
    Author provided the following answers.
    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].
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  6. #6
    Senior Member Vinod's Avatar
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    Re: Mean value theorem

    Quote Originally Posted by SlipEternal View Post
    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,
    I understood answer to first question.But please provide some explanation to the answer of second question.
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  7. #7
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    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, $h'(c) = 0$.
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  8. #8
    Senior Member Vinod's Avatar
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    Re: Mean value theorem

    Quote Originally Posted by SlipEternal View Post
    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.
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  9. #9
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    Re: Mean value theorem

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

    Example:

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

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

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

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