Using the M.V.T. to prove certain properties of a function
$begingroup$
Suppose that $f'(x) geq 10$ for all $x$ in $[0,5]$. Find $a>0$, as large as possible, such that there is guaranteed to be an interval $I$ of length $a$ contained in $[0,5]$ such that $|f|geq 2$ on $I$. Justify your answer using the mean value theorem.
My professor gave us the solution, however, I am looking for more intuition on how it works. Let $b<d$ be in the interval $[0,5]$, then there is $c$ between $b$ and $d$ such that $f'(c) = frac{f(b)-f(d)}{b-d}$
Use $f'geq 10$, so $frac{f(b)-f(d)}{b-d}geq 10$
Now choose $f(b) = -2$, $f(d)=2$, then $frac{-4}{b-d}geq10$
length = $d-bleqfrac{2}{5}$, use $I$ to be longer piece of $(0,b)$ and $(d,5)$. He also asked us to explain why it still works even if a $b,d$ such that $f(b)=-2$,$f(d)=2$.
I am confused on the choice of $f(b)$ and $f(d)$. Why did we choose those precise values? Is it because it is the smallest allowed value that $f$ can be on $I$? I am just looking for more intuition on how this problem works in general. Also I am guessing it still works if $f(b)neq-2$ and $f(d)neq2$ as we can just can take a bigger interval for $I$?
real-analysis proof-explanation
asked Dec 14 '18 at 3:53
hkj447hkj447
636
$endgroup$
add a comment |
$begingroup$
Suppose that $f'(x) geq 10$ for all $x$ in $[0,5]$. Find $a>0$, as large as possible, such that there is guaranteed to be an interval $I$ of length $a$ contained in $[0,5]$ such that $|f|geq 2$ on $I$. Justify your answer using the mean value theorem.
My professor gave us the solution, however, I am looking for more intuition on how it works. Let $b<d$ be in the interval $[0,5]$, then there is $c$ between $b$ and $d$ such that $f'(c) = frac{f(b)-f(d)}{b-d}$
Use $f'geq 10$, so $frac{f(b)-f(d)}{b-d}geq 10$
Now choose $f(b) = -2$, $f(d)=2$, then $frac{-4}{b-d}geq10$
length = $d-bleqfrac{2}{5}$, use $I$ to be longer piece of $(0,b)$ and $(d,5)$. He also asked us to explain why it still works even if a $b,d$ such that $f(b)=-2$,$f(d)=2$.
I am confused on the choice of $f(b)$ and $f(d)$. Why did we choose those precise values? Is it because it is the smallest allowed value that $f$ can be on $I$? I am just looking for more intuition on how this problem works in general. Also I am guessing it still works if $f(b)neq-2$ and $f(d)neq2$ as we can just can take a bigger interval for $I$?
real-analysis proof-explanation
asked Dec 14 '18 at 3:53
hkj447hkj447
636
$endgroup$
add a comment |
$begingroup$
Suppose that $f'(x) geq 10$ for all $x$ in $[0,5]$. Find $a>0$, as large as possible, such that there is guaranteed to be an interval $I$ of length $a$ contained in $[0,5]$ such that $|f|geq 2$ on $I$. Justify your answer using the mean value theorem.
My professor gave us the solution, however, I am looking for more intuition on how it works. Let $b<d$ be in the interval $[0,5]$, then there is $c$ between $b$ and $d$ such that $f'(c) = frac{f(b)-f(d)}{b-d}$
Use $f'geq 10$, so $frac{f(b)-f(d)}{b-d}geq 10$
Now choose $f(b) = -2$, $f(d)=2$, then $frac{-4}{b-d}geq10$
length = $d-bleqfrac{2}{5}$, use $I$ to be longer piece of $(0,b)$ and $(d,5)$. He also asked us to explain why it still works even if a $b,d$ such that $f(b)=-2$,$f(d)=2$.
I am confused on the choice of $f(b)$ and $f(d)$. Why did we choose those precise values? Is it because it is the smallest allowed value that $f$ can be on $I$? I am just looking for more intuition on how this problem works in general. Also I am guessing it still works if $f(b)neq-2$ and $f(d)neq2$ as we can just can take a bigger interval for $I$?
real-analysis proof-explanation
asked Dec 14 '18 at 3:53
hkj447hkj447
636
$endgroup$
Suppose that $f'(x) geq 10$ for all $x$ in $[0,5]$. Find $a>0$, as large as possible, such that there is guaranteed to be an interval $I$ of length $a$ contained in $[0,5]$ such that $|f|geq 2$ on $I$. Justify your answer using the mean value theorem.
My professor gave us the solution, however, I am looking for more intuition on how it works. Let $b<d$ be in the interval $[0,5]$, then there is $c$ between $b$ and $d$ such that $f'(c) = frac{f(b)-f(d)}{b-d}$
Use $f'geq 10$, so $frac{f(b)-f(d)}{b-d}geq 10$
Now choose $f(b) = -2$, $f(d)=2$, then $frac{-4}{b-d}geq10$
length = $d-bleqfrac{2}{5}$, use $I$ to be longer piece of $(0,b)$ and $(d,5)$. He also asked us to explain why it still works even if a $b,d$ such that $f(b)=-2$,$f(d)=2$.
I am confused on the choice of $f(b)$ and $f(d)$. Why did we choose those precise values? Is it because it is the smallest allowed value that $f$ can be on $I$? I am just looking for more intuition on how this problem works in general. Also I am guessing it still works if $f(b)neq-2$ and $f(d)neq2$ as we can just can take a bigger interval for $I$?
real-analysis proof-explanation
real-analysis proof-explanation
asked Dec 14 '18 at 3:53
hkj447hkj447
636
asked Dec 14 '18 at 3:53
hkj447hkj447
636
edited Dec 14 '18 at 4:03
hkj447
asked Dec 14 '18 at 3:53
hkj447hkj447
636
asked Dec 14 '18 at 3:53
hkj447hkj447
636
asked Dec 14 '18 at 3:53
hkj447hkj447
636
636
add a comment |
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