Proving continuity of an integral $2$ [closed]
How do i prove that $f(x)$ is continuous for every $xgeq0$
I really tried to prove it with epsilon delta but failed i really need help thanks for advance.
$$f(x)=begin{cases}intlimits_a^b f(t) &xgeq2\f(x)=2 &x<2end{cases}$$
where $a = dfrac{x}{2} $ and $b=left(dfrac{x}{2}right )^2$
integration
asked Nov 26 at 17:31
Razi Awad
124
closed as unclear what you're asking by Sean Roberson, Leucippus, Shailesh, Guacho Perez, Cesareo Nov 27 at 2:18
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
How do i prove that $f(x)$ is continuous for every $xgeq0$
I really tried to prove it with epsilon delta but failed i really need help thanks for advance.
$$f(x)=begin{cases}intlimits_a^b f(t) &xgeq2\f(x)=2 &x<2end{cases}$$
where $a = dfrac{x}{2} $ and $b=left(dfrac{x}{2}right )^2$
integration
asked Nov 26 at 17:31
Razi Awad
124
closed as unclear what you're asking by Sean Roberson, Leucippus, Shailesh, Guacho Perez, Cesareo Nov 27 at 2:18
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Please make your question more clear. This is very poorly written.
– Sean Roberson
Nov 26 at 17:36
As noted in my post, there is a unique function $f$ that satifies this definition. That function is continuous everywhere, and differentiable everywhere except at $x = 2$. This was put on hold as unclear prematurely.
– Paul Sinclair
Nov 27 at 3:38
add a comment |
How do i prove that $f(x)$ is continuous for every $xgeq0$
I really tried to prove it with epsilon delta but failed i really need help thanks for advance.
$$f(x)=begin{cases}intlimits_a^b f(t) &xgeq2\f(x)=2 &x<2end{cases}$$
where $a = dfrac{x}{2} $ and $b=left(dfrac{x}{2}right )^2$
integration
asked Nov 26 at 17:31
Razi Awad
124
How do i prove that $f(x)$ is continuous for every $xgeq0$
I really tried to prove it with epsilon delta but failed i really need help thanks for advance.
$$f(x)=begin{cases}intlimits_a^b f(t) &xgeq2\f(x)=2 &x<2end{cases}$$
where $a = dfrac{x}{2} $ and $b=left(dfrac{x}{2}right )^2$
integration
integration
asked Nov 26 at 17:31
Razi Awad
124
asked Nov 26 at 17:31
Razi Awad
124
edited Nov 26 at 18:00
asked Nov 26 at 17:31
Razi Awad
124
asked Nov 26 at 17:31
Razi Awad
124
asked Nov 26 at 17:31
Razi Awad
124
124
closed as unclear what you're asking by Sean Roberson, Leucippus, Shailesh, Guacho Perez, Cesareo Nov 27 at 2:18
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Sean Roberson, Leucippus, Shailesh, Guacho Perez, Cesareo Nov 27 at 2:18
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Please make your question more clear. This is very poorly written.
– Sean Roberson
Nov 26 at 17:36
As noted in my post, there is a unique function $f$ that satifies this definition. That function is continuous everywhere, and differentiable everywhere except at $x = 2$. This was put on hold as unclear prematurely.
– Paul Sinclair
Nov 27 at 3:38
add a comment |
Please make your question more clear. This is very poorly written.
– Sean Roberson
Nov 26 at 17:36
As noted in my post, there is a unique function $f$ that satifies this definition. That function is continuous everywhere, and differentiable everywhere except at $x = 2$. This was put on hold as unclear prematurely.
– Paul Sinclair
Nov 27 at 3:38
Please make your question more clear. This is very poorly written.
– Sean Roberson
Nov 26 at 17:36
Please make your question more clear. This is very poorly written.
– Sean Roberson
Nov 26 at 17:36
As noted in my post, there is a unique function $f$ that satifies this definition. That function is continuous everywhere, and differentiable everywhere except at $x = 2$. This was put on hold as unclear prematurely.
– Paul Sinclair
Nov 27 at 3:38
As noted in my post, there is a unique function $f$ that satifies this definition. That function is continuous everywhere, and differentiable everywhere except at $x = 2$. This was put on hold as unclear prematurely.
– Paul Sinclair
Nov 27 at 3:38
add a comment |
1 Answer
1
active
oldest
votes
To solve this, you need to closely examine the definition of $f$. There is a unique function $f$ that satisfies these conditions.
We already know that $f(x) = 2$ on $(-infty, 2]$. On $(2,2sqrt 2]$, we have that $a(x) = x/2$ is less than $b(x) = x^2/4 le 2$. So the integral is over the region where $f = 2$, and thus $$f(x) = frac {x^2}2 - x$$
For $x in (2sqrt 2, 4]$, the integral is always over a region to the left of $x$, and thus already defined.
For $x > 4$, we know that $f$ satisfies an integral with smoothly varying endpoints, and thus should be differentiable. Letting $I(a,b) = int_a^b f(t),dt$, we get $$begin{align}frac{df}{dx} &= frac{partial I}{partial b}frac{db}{dx} + frac{partial I}{partial a}frac{da}{dx}\ &= fleft(frac {x^2}4right)frac x2 - fleft(frac x2right)frac 12end{align}$$
Rearranging $$fleft(frac {x^2}4right) = frac 2xleft[f'(x) - frac 12fleft(frac x2right)right]$$
Making a substitution,
$$f(t) = frac 1{sqrt t}left[f'(2sqrt t) - frac 12f(sqrt t)right]$$
Since for $t > 4$, we have $t > 2sqrt t > sqrt t$, this defines the value of $f$ at $t$ in terms of the value of the function and derivative at points to the left where it was already defined. By an argument similar to inductive definition, this defines $f$ everywhere.
answered Nov 27 at 0:22
Paul Sinclair
19.2k21441
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
To solve this, you need to closely examine the definition of $f$. There is a unique function $f$ that satisfies these conditions.
We already know that $f(x) = 2$ on $(-infty, 2]$. On $(2,2sqrt 2]$, we have that $a(x) = x/2$ is less than $b(x) = x^2/4 le 2$. So the integral is over the region where $f = 2$, and thus $$f(x) = frac {x^2}2 - x$$
For $x in (2sqrt 2, 4]$, the integral is always over a region to the left of $x$, and thus already defined.
For $x > 4$, we know that $f$ satisfies an integral with smoothly varying endpoints, and thus should be differentiable. Letting $I(a,b) = int_a^b f(t),dt$, we get $$begin{align}frac{df}{dx} &= frac{partial I}{partial b}frac{db}{dx} + frac{partial I}{partial a}frac{da}{dx}\ &= fleft(frac {x^2}4right)frac x2 - fleft(frac x2right)frac 12end{align}$$
Rearranging $$fleft(frac {x^2}4right) = frac 2xleft[f'(x) - frac 12fleft(frac x2right)right]$$
Making a substitution,
$$f(t) = frac 1{sqrt t}left[f'(2sqrt t) - frac 12f(sqrt t)right]$$
Since for $t > 4$, we have $t > 2sqrt t > sqrt t$, this defines the value of $f$ at $t$ in terms of the value of the function and derivative at points to the left where it was already defined. By an argument similar to inductive definition, this defines $f$ everywhere.
answered Nov 27 at 0:22
Paul Sinclair
19.2k21441
add a comment |
To solve this, you need to closely examine the definition of $f$. There is a unique function $f$ that satisfies these conditions.
We already know that $f(x) = 2$ on $(-infty, 2]$. On $(2,2sqrt 2]$, we have that $a(x) = x/2$ is less than $b(x) = x^2/4 le 2$. So the integral is over the region where $f = 2$, and thus $$f(x) = frac {x^2}2 - x$$
For $x in (2sqrt 2, 4]$, the integral is always over a region to the left of $x$, and thus already defined.
For $x > 4$, we know that $f$ satisfies an integral with smoothly varying endpoints, and thus should be differentiable. Letting $I(a,b) = int_a^b f(t),dt$, we get $$begin{align}frac{df}{dx} &= frac{partial I}{partial b}frac{db}{dx} + frac{partial I}{partial a}frac{da}{dx}\ &= fleft(frac {x^2}4right)frac x2 - fleft(frac x2right)frac 12end{align}$$
Rearranging $$fleft(frac {x^2}4right) = frac 2xleft[f'(x) - frac 12fleft(frac x2right)right]$$
Making a substitution,
$$f(t) = frac 1{sqrt t}left[f'(2sqrt t) - frac 12f(sqrt t)right]$$
Since for $t > 4$, we have $t > 2sqrt t > sqrt t$, this defines the value of $f$ at $t$ in terms of the value of the function and derivative at points to the left where it was already defined. By an argument similar to inductive definition, this defines $f$ everywhere.
answered Nov 27 at 0:22
Paul Sinclair
19.2k21441
add a comment |
To solve this, you need to closely examine the definition of $f$. There is a unique function $f$ that satisfies these conditions.
We already know that $f(x) = 2$ on $(-infty, 2]$. On $(2,2sqrt 2]$, we have that $a(x) = x/2$ is less than $b(x) = x^2/4 le 2$. So the integral is over the region where $f = 2$, and thus $$f(x) = frac {x^2}2 - x$$
For $x in (2sqrt 2, 4]$, the integral is always over a region to the left of $x$, and thus already defined.
For $x > 4$, we know that $f$ satisfies an integral with smoothly varying endpoints, and thus should be differentiable. Letting $I(a,b) = int_a^b f(t),dt$, we get $$begin{align}frac{df}{dx} &= frac{partial I}{partial b}frac{db}{dx} + frac{partial I}{partial a}frac{da}{dx}\ &= fleft(frac {x^2}4right)frac x2 - fleft(frac x2right)frac 12end{align}$$
Rearranging $$fleft(frac {x^2}4right) = frac 2xleft[f'(x) - frac 12fleft(frac x2right)right]$$
Making a substitution,
$$f(t) = frac 1{sqrt t}left[f'(2sqrt t) - frac 12f(sqrt t)right]$$
Since for $t > 4$, we have $t > 2sqrt t > sqrt t$, this defines the value of $f$ at $t$ in terms of the value of the function and derivative at points to the left where it was already defined. By an argument similar to inductive definition, this defines $f$ everywhere.
answered Nov 27 at 0:22
Paul Sinclair
19.2k21441
To solve this, you need to closely examine the definition of $f$. There is a unique function $f$ that satisfies these conditions.
We already know that $f(x) = 2$ on $(-infty, 2]$. On $(2,2sqrt 2]$, we have that $a(x) = x/2$ is less than $b(x) = x^2/4 le 2$. So the integral is over the region where $f = 2$, and thus $$f(x) = frac {x^2}2 - x$$
For $x in (2sqrt 2, 4]$, the integral is always over a region to the left of $x$, and thus already defined.
For $x > 4$, we know that $f$ satisfies an integral with smoothly varying endpoints, and thus should be differentiable. Letting $I(a,b) = int_a^b f(t),dt$, we get $$begin{align}frac{df}{dx} &= frac{partial I}{partial b}frac{db}{dx} + frac{partial I}{partial a}frac{da}{dx}\ &= fleft(frac {x^2}4right)frac x2 - fleft(frac x2right)frac 12end{align}$$
Rearranging $$fleft(frac {x^2}4right) = frac 2xleft[f'(x) - frac 12fleft(frac x2right)right]$$
Making a substitution,
$$f(t) = frac 1{sqrt t}left[f'(2sqrt t) - frac 12f(sqrt t)right]$$
Since for $t > 4$, we have $t > 2sqrt t > sqrt t$, this defines the value of $f$ at $t$ in terms of the value of the function and derivative at points to the left where it was already defined. By an argument similar to inductive definition, this defines $f$ everywhere.
answered Nov 27 at 0:22
Paul Sinclair
19.2k21441
edited Nov 27 at 3:41
answered Nov 27 at 0:22
Paul Sinclair
19.2k21441
answered Nov 27 at 0:22
Paul Sinclair
19.2k21441
answered Nov 27 at 0:22
Paul Sinclair
19.2k21441
19.2k21441
add a comment |
add a comment |
Please make your question more clear. This is very poorly written.
– Sean Roberson
Nov 26 at 17:36
As noted in my post, there is a unique function $f$ that satifies this definition. That function is continuous everywhere, and differentiable everywhere except at $x = 2$. This was put on hold as unclear prematurely.
– Paul Sinclair
Nov 27 at 3:38