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July 23rd, 2018, 04:06 AM   #1
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Can someone solve this problem?

this is not a homework.

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July 23rd, 2018, 05:10 AM   #2
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did i solve for A) correctly?

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July 23rd, 2018, 07:12 AM   #3
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did i solve for A) correctly?

not quite ...

note $\displaystyle \int_2^3 f(t) \, dt = -0.5
$

for part (b), note $g'(x) = f(x)$

look for the intervals where $g'(x) > 0$
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July 23rd, 2018, 08:04 AM   #4
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Are you saying that function g(x) is not increasing for interval 2 < x < 3?

Last edited by skipjack; July 23rd, 2018 at 11:18 AM.
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July 23rd, 2018, 09:26 AM   #5
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Are you saying that function g(x) is not increasing for interval 2 < x < 3?
That is correct ...

note on $2 < x < 3$, $g'(x) = f(x) < 0 \implies g(x)$ is decreasing.
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Last edited by skipjack; July 23rd, 2018 at 11:18 AM.
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July 23rd, 2018, 09:40 AM   #6
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But at that interval for higher value of $x$ we get higher value of $f(x)$, but never mind.
Can you just do for D) please? That would be great.

Last edited by skipjack; July 23rd, 2018 at 10:32 AM.
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July 23rd, 2018, 10:47 AM   #7
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But at that interval for higher value of $x$ we get higher value of $f(x)$, but never mind.
Can you just do for D) please? That would be great.
Do you see that $g'(x) = f(x)$? If so then you can compute $h'(2)$ using the product rule:
\[ h'(2) = 2f(2) + g(2) \]
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July 23rd, 2018, 11:23 AM   #8
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What are your answers for (B) and (C), 1ucid?
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July 23rd, 2018, 11:29 AM   #9
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is this ok?

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July 23rd, 2018, 11:41 AM   #10
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What are your answers for (B) and (C), 1ucid?
This is for (B), but for (C) I'm not sure what is the right answer.


Last edited by skipjack; July 23rd, 2018 at 12:06 PM.
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