Mixing
Proving a statement with the mean value theorem
$begingroup$
If $f:[-2,2]rightarrow R$ is a continuous function, there exists $c $ in $(-2,2)$ such that $$int_{-2}^cf(x)dx=left( frac{4}{c^3}-frac{c}{4}right) f(c)$$.I tried to prove this using the mean value theorem.
So there exists $c$ in $(-2,2)$ such that $$int_{-2}^2f(x)dx=4f(c)$$.
Can somebody tell me what should I do next, please?
calculus
asked Jan 11 at 11:53
GaboruGaboru
827
$endgroup$
add a comment |
$begingroup$
If $f:[-2,2]rightarrow R$ is a continuous function, there exists $c $ in $(-2,2)$ such that $$int_{-2}^cf(x)dx=left( frac{4}{c^3}-frac{c}{4}right) f(c)$$.I tried to prove this using the mean value theorem.
So there exists $c$ in $(-2,2)$ such that $$int_{-2}^2f(x)dx=4f(c)$$.
Can somebody tell me what should I do next, please?
calculus
asked Jan 11 at 11:53
GaboruGaboru
827
$endgroup$
$begingroup$
i edited the post
$endgroup$
– Gaboru
Jan 11 at 12:09
$begingroup$
Seen the edit, it is ok now.Try to apply the MVT with a modified function.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 12:09
add a comment |
$begingroup$
If $f:[-2,2]rightarrow R$ is a continuous function, there exists $c $ in $(-2,2)$ such that $$int_{-2}^cf(x)dx=left( frac{4}{c^3}-frac{c}{4}right) f(c)$$.I tried to prove this using the mean value theorem.
So there exists $c$ in $(-2,2)$ such that $$int_{-2}^2f(x)dx=4f(c)$$.
Can somebody tell me what should I do next, please?
calculus
asked Jan 11 at 11:53
GaboruGaboru
827
$endgroup$
If $f:[-2,2]rightarrow R$ is a continuous function, there exists $c $ in $(-2,2)$ such that $$int_{-2}^cf(x)dx=left( frac{4}{c^3}-frac{c}{4}right) f(c)$$.I tried to prove this using the mean value theorem.
So there exists $c$ in $(-2,2)$ such that $$int_{-2}^2f(x)dx=4f(c)$$.
Can somebody tell me what should I do next, please?
calculus
calculus
asked Jan 11 at 11:53
GaboruGaboru
827
asked Jan 11 at 11:53
GaboruGaboru
827
edited Jan 11 at 12:08
Gaboru
asked Jan 11 at 11:53
GaboruGaboru
827
asked Jan 11 at 11:53
GaboruGaboru
827
asked Jan 11 at 11:53
GaboruGaboru
827
827
$begingroup$
i edited the post
$endgroup$
– Gaboru
Jan 11 at 12:09
$begingroup$
Seen the edit, it is ok now.Try to apply the MVT with a modified function.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 12:09
add a comment |
$begingroup$
i edited the post
$endgroup$
– Gaboru
Jan 11 at 12:09
$begingroup$
Seen the edit, it is ok now.Try to apply the MVT with a modified function.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 12:09
$begingroup$
i edited the post
$endgroup$
– Gaboru
Jan 11 at 12:09
$begingroup$
i edited the post
$endgroup$
– Gaboru
Jan 11 at 12:09
$begingroup$
Seen the edit, it is ok now.Try to apply the MVT with a modified function.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 12:09
$begingroup$
Seen the edit, it is ok now.Try to apply the MVT with a modified function.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 12:09
add a comment |
1 Answer
1
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$begingroup$
First, we define the function $g : [-2,2] to mathbb R$ by $g(x) = int_{-2}^x f(t)mbox{dt}$. By the fundamental theorem of calculus, $g$ is differentiable, with $g'(x) = f(x)$.
Consider the function $h(x) = 16 - x^4$ on $[-2,2]$. Then, note that:
$$
frac{4}{c^3}- frac c4 = frac{16 - c^4}{4c^3} = -frac{h(c)}{h'(c)}
$$
We are essentially asked to show that there is a $c$ such that $g(c) = frac{-f(c)h(c)}{h'(c)}$, or $h'(c)g(c) = -g'(c)h(c)$, or $(gh)'(c) = 0$ by the product rule.
Note that $h(-2) = h(2) = 0$. Conclude that $gh(-2) = gh(2) = 0$. Can you use some theorem now?
answered Jan 11 at 12:16
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
$endgroup$
$begingroup$
thank you! I did it using Rolle theorem
$endgroup$
– Gaboru
Jan 11 at 13:42
$begingroup$
That is exactly what I expected! Remember, identifying tricks is key to solving questions. Keep this one in mind.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 14:17
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
First, we define the function $g : [-2,2] to mathbb R$ by $g(x) = int_{-2}^x f(t)mbox{dt}$. By the fundamental theorem of calculus, $g$ is differentiable, with $g'(x) = f(x)$.
Consider the function $h(x) = 16 - x^4$ on $[-2,2]$. Then, note that:
$$
frac{4}{c^3}- frac c4 = frac{16 - c^4}{4c^3} = -frac{h(c)}{h'(c)}
$$
We are essentially asked to show that there is a $c$ such that $g(c) = frac{-f(c)h(c)}{h'(c)}$, or $h'(c)g(c) = -g'(c)h(c)$, or $(gh)'(c) = 0$ by the product rule.
Note that $h(-2) = h(2) = 0$. Conclude that $gh(-2) = gh(2) = 0$. Can you use some theorem now?
answered Jan 11 at 12:16
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
$endgroup$
$begingroup$
thank you! I did it using Rolle theorem
$endgroup$
– Gaboru
Jan 11 at 13:42
$begingroup$
That is exactly what I expected! Remember, identifying tricks is key to solving questions. Keep this one in mind.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 14:17
add a comment |
$begingroup$
First, we define the function $g : [-2,2] to mathbb R$ by $g(x) = int_{-2}^x f(t)mbox{dt}$. By the fundamental theorem of calculus, $g$ is differentiable, with $g'(x) = f(x)$.
Consider the function $h(x) = 16 - x^4$ on $[-2,2]$. Then, note that:
$$
frac{4}{c^3}- frac c4 = frac{16 - c^4}{4c^3} = -frac{h(c)}{h'(c)}
$$
We are essentially asked to show that there is a $c$ such that $g(c) = frac{-f(c)h(c)}{h'(c)}$, or $h'(c)g(c) = -g'(c)h(c)$, or $(gh)'(c) = 0$ by the product rule.
Note that $h(-2) = h(2) = 0$. Conclude that $gh(-2) = gh(2) = 0$. Can you use some theorem now?
answered Jan 11 at 12:16
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
$endgroup$
$begingroup$
thank you! I did it using Rolle theorem
$endgroup$
– Gaboru
Jan 11 at 13:42
$begingroup$
That is exactly what I expected! Remember, identifying tricks is key to solving questions. Keep this one in mind.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 14:17
add a comment |
$begingroup$
First, we define the function $g : [-2,2] to mathbb R$ by $g(x) = int_{-2}^x f(t)mbox{dt}$. By the fundamental theorem of calculus, $g$ is differentiable, with $g'(x) = f(x)$.
Consider the function $h(x) = 16 - x^4$ on $[-2,2]$. Then, note that:
$$
frac{4}{c^3}- frac c4 = frac{16 - c^4}{4c^3} = -frac{h(c)}{h'(c)}
$$
We are essentially asked to show that there is a $c$ such that $g(c) = frac{-f(c)h(c)}{h'(c)}$, or $h'(c)g(c) = -g'(c)h(c)$, or $(gh)'(c) = 0$ by the product rule.
Note that $h(-2) = h(2) = 0$. Conclude that $gh(-2) = gh(2) = 0$. Can you use some theorem now?
answered Jan 11 at 12:16
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
$endgroup$
First, we define the function $g : [-2,2] to mathbb R$ by $g(x) = int_{-2}^x f(t)mbox{dt}$. By the fundamental theorem of calculus, $g$ is differentiable, with $g'(x) = f(x)$.
Consider the function $h(x) = 16 - x^4$ on $[-2,2]$. Then, note that:
$$
frac{4}{c^3}- frac c4 = frac{16 - c^4}{4c^3} = -frac{h(c)}{h'(c)}
$$
We are essentially asked to show that there is a $c$ such that $g(c) = frac{-f(c)h(c)}{h'(c)}$, or $h'(c)g(c) = -g'(c)h(c)$, or $(gh)'(c) = 0$ by the product rule.
Note that $h(-2) = h(2) = 0$. Conclude that $gh(-2) = gh(2) = 0$. Can you use some theorem now?
answered Jan 11 at 12:16
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
answered Jan 11 at 12:16
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
answered Jan 11 at 12:16
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
answered Jan 11 at 12:16
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
38.3k33376
$begingroup$
thank you! I did it using Rolle theorem
$endgroup$
– Gaboru
Jan 11 at 13:42
$begingroup$
That is exactly what I expected! Remember, identifying tricks is key to solving questions. Keep this one in mind.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 14:17
add a comment |
$begingroup$
thank you! I did it using Rolle theorem
$endgroup$
– Gaboru
Jan 11 at 13:42
$begingroup$
That is exactly what I expected! Remember, identifying tricks is key to solving questions. Keep this one in mind.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 14:17
$begingroup$
thank you! I did it using Rolle theorem
$endgroup$
– Gaboru
Jan 11 at 13:42
$begingroup$
thank you! I did it using Rolle theorem
$endgroup$
– Gaboru
Jan 11 at 13:42
$begingroup$
That is exactly what I expected! Remember, identifying tricks is key to solving questions. Keep this one in mind.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 14:17
$begingroup$
That is exactly what I expected! Remember, identifying tricks is key to solving questions. Keep this one in mind.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 14:17
add a comment |
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$begingroup$
i edited the post
$endgroup$
– Gaboru
Jan 11 at 12:09
$begingroup$
Seen the edit, it is ok now.Try to apply the MVT with a modified function.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 12:09