one interesting calculus problem












0















let $f(x)=sin(frac{pi}{2}x)$
and $g(x)$ satisfies $0leq g(x)leq 1$ in $xin [0,2]$ then $h(x)=f(x+g(x))$ is part of quadratic function.
find the maximum value of $int_0^2 vert h'(x) vert dx$




This is high school calculus problem. But I don't know where to start.






calculus integration







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  • 1




    What do you mean by "part of quadratic function"?
    – J.G.
    Jan 2 at 19:25












  • I think the question is just telling that h(x) is quadratic in [0,2]
    – scitamehtam
    Jan 2 at 19:34
















0















let $f(x)=sin(frac{pi}{2}x)$
and $g(x)$ satisfies $0leq g(x)leq 1$ in $xin [0,2]$ then $h(x)=f(x+g(x))$ is part of quadratic function.
find the maximum value of $int_0^2 vert h'(x) vert dx$




This is high school calculus problem. But I don't know where to start.






calculus integration







share|cite|improve this question


















  • 1




    What do you mean by "part of quadratic function"?
    – J.G.
    Jan 2 at 19:25












  • I think the question is just telling that h(x) is quadratic in [0,2]
    – scitamehtam
    Jan 2 at 19:34














0












0








0








let $f(x)=sin(frac{pi}{2}x)$
and $g(x)$ satisfies $0leq g(x)leq 1$ in $xin [0,2]$ then $h(x)=f(x+g(x))$ is part of quadratic function.
find the maximum value of $int_0^2 vert h'(x) vert dx$




This is high school calculus problem. But I don't know where to start.






calculus integration







share|cite|improve this question














let $f(x)=sin(frac{pi}{2}x)$
and $g(x)$ satisfies $0leq g(x)leq 1$ in $xin [0,2]$ then $h(x)=f(x+g(x))$ is part of quadratic function.
find the maximum value of $int_0^2 vert h'(x) vert dx$




This is high school calculus problem. But I don't know where to start.






calculus integration




calculus integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 19:17









scitamehtamscitamehtam

334




334








  • 1




    What do you mean by "part of quadratic function"?
    – J.G.
    Jan 2 at 19:25












  • I think the question is just telling that h(x) is quadratic in [0,2]
    – scitamehtam
    Jan 2 at 19:34














  • 1




    What do you mean by "part of quadratic function"?
    – J.G.
    Jan 2 at 19:25












  • I think the question is just telling that h(x) is quadratic in [0,2]
    – scitamehtam
    Jan 2 at 19:34








1




1




What do you mean by "part of quadratic function"?
– J.G.
Jan 2 at 19:25






What do you mean by "part of quadratic function"?
– J.G.
Jan 2 at 19:25














I think the question is just telling that h(x) is quadratic in [0,2]
– scitamehtam
Jan 2 at 19:34




I think the question is just telling that h(x) is quadratic in [0,2]
– scitamehtam
Jan 2 at 19:34










1 Answer
1






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oldest

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0














The question doesn't state it explicitly, but for $h'left(xright)$ to exist, $gleft(xright)$ must be differentiable. Also, as asked by J.G., I'm also not clear what "part of quadratic function" is supposed to mean. However, I believe it likely is important because, without any other restrictions, there is no maximum of $int_{0}^{2} lvert h'left(xright) rvert dx$. This is due to the possibility of $hleft(xright)$ increasing and decreasing by a distance close to $1$ arbitrarily often in a very short span.



To see this, consider that in sections, say from $a$ to $b$, where $h'left(xright) ge 0$, we get



$$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} h'left(xright) dx = hleft(bright) - hleft(aright) tag{1}label{eq1} $$



In other words, it is the end less the start value of $h$, which works out to the highest less lowest values. For other sections, say from $c$ to $d$, where $h'left(xright) lt 0$, we instead get



$$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} -h'left(xright) dx = hleft(aright) - hleft(bright) tag{2}label{eq2} $$



As $h$ is decreasing, this is once again the highest less lowest values. Now, let's say at $x = 0$ that $gleft(0right) = 1$ so $hleft(0right) = sin frac{pi}{2} = 1$. Now, if $gleft(xright)$ were to very quickly go to $0$, say within $10^{-1000}$, then $hleft(xright)$ would decrease very quickly to close to $sinleft(0right) = 0$. The integral over this portion would then be very close to $1$. Next, $gleft(xright)$ could then very quickly go back to $1$, say again within $10^{-1000}$, thus increasing $hleft(xright)$ very quickly to close to $1$ again, once again adding close to $1$ to the integral. This can be done an arbitrary number of times within any minimal span you care to choose. As each integral section value being added is always positive, this means there is no maximum value.



I'm sorry this explanation is particularly "clean" mathematically, as I thought that using things like $delta$ and $epsilon$ would perhaps confuse the situation. I trust that it's clear enough you get the idea and that it can perhaps help you solve the question yourself based on what "part of quadratic function" means.






share|cite|improve this answer























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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

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    0














    The question doesn't state it explicitly, but for $h'left(xright)$ to exist, $gleft(xright)$ must be differentiable. Also, as asked by J.G., I'm also not clear what "part of quadratic function" is supposed to mean. However, I believe it likely is important because, without any other restrictions, there is no maximum of $int_{0}^{2} lvert h'left(xright) rvert dx$. This is due to the possibility of $hleft(xright)$ increasing and decreasing by a distance close to $1$ arbitrarily often in a very short span.



    To see this, consider that in sections, say from $a$ to $b$, where $h'left(xright) ge 0$, we get



    $$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} h'left(xright) dx = hleft(bright) - hleft(aright) tag{1}label{eq1} $$



    In other words, it is the end less the start value of $h$, which works out to the highest less lowest values. For other sections, say from $c$ to $d$, where $h'left(xright) lt 0$, we instead get



    $$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} -h'left(xright) dx = hleft(aright) - hleft(bright) tag{2}label{eq2} $$



    As $h$ is decreasing, this is once again the highest less lowest values. Now, let's say at $x = 0$ that $gleft(0right) = 1$ so $hleft(0right) = sin frac{pi}{2} = 1$. Now, if $gleft(xright)$ were to very quickly go to $0$, say within $10^{-1000}$, then $hleft(xright)$ would decrease very quickly to close to $sinleft(0right) = 0$. The integral over this portion would then be very close to $1$. Next, $gleft(xright)$ could then very quickly go back to $1$, say again within $10^{-1000}$, thus increasing $hleft(xright)$ very quickly to close to $1$ again, once again adding close to $1$ to the integral. This can be done an arbitrary number of times within any minimal span you care to choose. As each integral section value being added is always positive, this means there is no maximum value.



    I'm sorry this explanation is particularly "clean" mathematically, as I thought that using things like $delta$ and $epsilon$ would perhaps confuse the situation. I trust that it's clear enough you get the idea and that it can perhaps help you solve the question yourself based on what "part of quadratic function" means.






    share|cite|improve this answer




























      0














      The question doesn't state it explicitly, but for $h'left(xright)$ to exist, $gleft(xright)$ must be differentiable. Also, as asked by J.G., I'm also not clear what "part of quadratic function" is supposed to mean. However, I believe it likely is important because, without any other restrictions, there is no maximum of $int_{0}^{2} lvert h'left(xright) rvert dx$. This is due to the possibility of $hleft(xright)$ increasing and decreasing by a distance close to $1$ arbitrarily often in a very short span.



      To see this, consider that in sections, say from $a$ to $b$, where $h'left(xright) ge 0$, we get



      $$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} h'left(xright) dx = hleft(bright) - hleft(aright) tag{1}label{eq1} $$



      In other words, it is the end less the start value of $h$, which works out to the highest less lowest values. For other sections, say from $c$ to $d$, where $h'left(xright) lt 0$, we instead get



      $$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} -h'left(xright) dx = hleft(aright) - hleft(bright) tag{2}label{eq2} $$



      As $h$ is decreasing, this is once again the highest less lowest values. Now, let's say at $x = 0$ that $gleft(0right) = 1$ so $hleft(0right) = sin frac{pi}{2} = 1$. Now, if $gleft(xright)$ were to very quickly go to $0$, say within $10^{-1000}$, then $hleft(xright)$ would decrease very quickly to close to $sinleft(0right) = 0$. The integral over this portion would then be very close to $1$. Next, $gleft(xright)$ could then very quickly go back to $1$, say again within $10^{-1000}$, thus increasing $hleft(xright)$ very quickly to close to $1$ again, once again adding close to $1$ to the integral. This can be done an arbitrary number of times within any minimal span you care to choose. As each integral section value being added is always positive, this means there is no maximum value.



      I'm sorry this explanation is particularly "clean" mathematically, as I thought that using things like $delta$ and $epsilon$ would perhaps confuse the situation. I trust that it's clear enough you get the idea and that it can perhaps help you solve the question yourself based on what "part of quadratic function" means.






      share|cite|improve this answer


























        0












        0








        0






        The question doesn't state it explicitly, but for $h'left(xright)$ to exist, $gleft(xright)$ must be differentiable. Also, as asked by J.G., I'm also not clear what "part of quadratic function" is supposed to mean. However, I believe it likely is important because, without any other restrictions, there is no maximum of $int_{0}^{2} lvert h'left(xright) rvert dx$. This is due to the possibility of $hleft(xright)$ increasing and decreasing by a distance close to $1$ arbitrarily often in a very short span.



        To see this, consider that in sections, say from $a$ to $b$, where $h'left(xright) ge 0$, we get



        $$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} h'left(xright) dx = hleft(bright) - hleft(aright) tag{1}label{eq1} $$



        In other words, it is the end less the start value of $h$, which works out to the highest less lowest values. For other sections, say from $c$ to $d$, where $h'left(xright) lt 0$, we instead get



        $$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} -h'left(xright) dx = hleft(aright) - hleft(bright) tag{2}label{eq2} $$



        As $h$ is decreasing, this is once again the highest less lowest values. Now, let's say at $x = 0$ that $gleft(0right) = 1$ so $hleft(0right) = sin frac{pi}{2} = 1$. Now, if $gleft(xright)$ were to very quickly go to $0$, say within $10^{-1000}$, then $hleft(xright)$ would decrease very quickly to close to $sinleft(0right) = 0$. The integral over this portion would then be very close to $1$. Next, $gleft(xright)$ could then very quickly go back to $1$, say again within $10^{-1000}$, thus increasing $hleft(xright)$ very quickly to close to $1$ again, once again adding close to $1$ to the integral. This can be done an arbitrary number of times within any minimal span you care to choose. As each integral section value being added is always positive, this means there is no maximum value.



        I'm sorry this explanation is particularly "clean" mathematically, as I thought that using things like $delta$ and $epsilon$ would perhaps confuse the situation. I trust that it's clear enough you get the idea and that it can perhaps help you solve the question yourself based on what "part of quadratic function" means.






        share|cite|improve this answer














        The question doesn't state it explicitly, but for $h'left(xright)$ to exist, $gleft(xright)$ must be differentiable. Also, as asked by J.G., I'm also not clear what "part of quadratic function" is supposed to mean. However, I believe it likely is important because, without any other restrictions, there is no maximum of $int_{0}^{2} lvert h'left(xright) rvert dx$. This is due to the possibility of $hleft(xright)$ increasing and decreasing by a distance close to $1$ arbitrarily often in a very short span.



        To see this, consider that in sections, say from $a$ to $b$, where $h'left(xright) ge 0$, we get



        $$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} h'left(xright) dx = hleft(bright) - hleft(aright) tag{1}label{eq1} $$



        In other words, it is the end less the start value of $h$, which works out to the highest less lowest values. For other sections, say from $c$ to $d$, where $h'left(xright) lt 0$, we instead get



        $$int_{a}^{b} lvert h'left(xright) rvert dx = int_{a}^{b} -h'left(xright) dx = hleft(aright) - hleft(bright) tag{2}label{eq2} $$



        As $h$ is decreasing, this is once again the highest less lowest values. Now, let's say at $x = 0$ that $gleft(0right) = 1$ so $hleft(0right) = sin frac{pi}{2} = 1$. Now, if $gleft(xright)$ were to very quickly go to $0$, say within $10^{-1000}$, then $hleft(xright)$ would decrease very quickly to close to $sinleft(0right) = 0$. The integral over this portion would then be very close to $1$. Next, $gleft(xright)$ could then very quickly go back to $1$, say again within $10^{-1000}$, thus increasing $hleft(xright)$ very quickly to close to $1$ again, once again adding close to $1$ to the integral. This can be done an arbitrary number of times within any minimal span you care to choose. As each integral section value being added is always positive, this means there is no maximum value.



        I'm sorry this explanation is particularly "clean" mathematically, as I thought that using things like $delta$ and $epsilon$ would perhaps confuse the situation. I trust that it's clear enough you get the idea and that it can perhaps help you solve the question yourself based on what "part of quadratic function" means.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 6:41

























        answered Jan 4 at 6:31









        John OmielanJohn Omielan

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