Why do we include the opposite value in the circle unit of a critical number if it falls in the domain?
f(t) = 9t + 9cot(t/2). Find the absolute minimum and maximum values in the domain [π/4, 7π/4]
the derivative of (df/dt)f(t) = 9+(-9csc^2(t/2)*1/2).
When we set the derivative of f(t)=0, then:
t = π/2
Now I understand 3π/2 falls in the domain however, why do we include it as a critical value of x? It cannot simply be because it runs parallel to the critical value of π/2 otherwise, what is the import or significance of it being parallel?
calculus trigonometry differential-geometry
asked Nov 20 at 6:08
Seungsoo Im
11
add a comment |
f(t) = 9t + 9cot(t/2). Find the absolute minimum and maximum values in the domain [π/4, 7π/4]
the derivative of (df/dt)f(t) = 9+(-9csc^2(t/2)*1/2).
When we set the derivative of f(t)=0, then:
t = π/2
Now I understand 3π/2 falls in the domain however, why do we include it as a critical value of x? It cannot simply be because it runs parallel to the critical value of π/2 otherwise, what is the import or significance of it being parallel?
calculus trigonometry differential-geometry
asked Nov 20 at 6:08
Seungsoo Im
11
Check your derivative.
– David G. Stork
Nov 20 at 6:58
its right although it can definitely be simplified further.
– Seungsoo Im
Nov 23 at 23:16
add a comment |
f(t) = 9t + 9cot(t/2). Find the absolute minimum and maximum values in the domain [π/4, 7π/4]
the derivative of (df/dt)f(t) = 9+(-9csc^2(t/2)*1/2).
When we set the derivative of f(t)=0, then:
t = π/2
Now I understand 3π/2 falls in the domain however, why do we include it as a critical value of x? It cannot simply be because it runs parallel to the critical value of π/2 otherwise, what is the import or significance of it being parallel?
calculus trigonometry differential-geometry
asked Nov 20 at 6:08
Seungsoo Im
11
f(t) = 9t + 9cot(t/2). Find the absolute minimum and maximum values in the domain [π/4, 7π/4]
the derivative of (df/dt)f(t) = 9+(-9csc^2(t/2)*1/2).
When we set the derivative of f(t)=0, then:
t = π/2
Now I understand 3π/2 falls in the domain however, why do we include it as a critical value of x? It cannot simply be because it runs parallel to the critical value of π/2 otherwise, what is the import or significance of it being parallel?
calculus trigonometry differential-geometry
calculus trigonometry differential-geometry
asked Nov 20 at 6:08
Seungsoo Im
11
asked Nov 20 at 6:08
Seungsoo Im
11
asked Nov 20 at 6:08
Seungsoo Im
11
asked Nov 20 at 6:08
Seungsoo Im
11
asked Nov 20 at 6:08
Seungsoo Im
11
11
Check your derivative.
– David G. Stork
Nov 20 at 6:58
its right although it can definitely be simplified further.
– Seungsoo Im
Nov 23 at 23:16
add a comment |
Check your derivative.
– David G. Stork
Nov 20 at 6:58
its right although it can definitely be simplified further.
– Seungsoo Im
Nov 23 at 23:16
Check your derivative.
– David G. Stork
Nov 20 at 6:58
Check your derivative.
– David G. Stork
Nov 20 at 6:58
its right although it can definitely be simplified further.
– Seungsoo Im
Nov 23 at 23:16
its right although it can definitely be simplified further.
– Seungsoo Im
Nov 23 at 23:16
add a comment |
1 Answer
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Here's a plot of your function (blue) and its derivative (orange):
Because the derivative is always positive, the minimum of your function is at the left (at $pi/4$) and the maximum is at the right (at $7 pi/4$).
$f(t) = 9 t + cot left( {t over 2} right)$
and
${partial over partial t} f(t) = 9 - {1 over 2} csc^2 left( {t over 2} right)$
answered Nov 20 at 7:03
David G. Stork
9,69621232
wait.. according to my answers, the minimum is at π/2 and the maximum is I believe 7π/4
– Seungsoo Im
Nov 23 at 23:18
nevermind, the maximum value is when t = 3π/2
– Seungsoo Im
Nov 24 at 0:35
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's a plot of your function (blue) and its derivative (orange):
Because the derivative is always positive, the minimum of your function is at the left (at $pi/4$) and the maximum is at the right (at $7 pi/4$).
$f(t) = 9 t + cot left( {t over 2} right)$
and
${partial over partial t} f(t) = 9 - {1 over 2} csc^2 left( {t over 2} right)$
answered Nov 20 at 7:03
David G. Stork
9,69621232
wait.. according to my answers, the minimum is at π/2 and the maximum is I believe 7π/4
– Seungsoo Im
Nov 23 at 23:18
nevermind, the maximum value is when t = 3π/2
– Seungsoo Im
Nov 24 at 0:35
add a comment |
Here's a plot of your function (blue) and its derivative (orange):
Because the derivative is always positive, the minimum of your function is at the left (at $pi/4$) and the maximum is at the right (at $7 pi/4$).
$f(t) = 9 t + cot left( {t over 2} right)$
and
${partial over partial t} f(t) = 9 - {1 over 2} csc^2 left( {t over 2} right)$
answered Nov 20 at 7:03
David G. Stork
9,69621232
wait.. according to my answers, the minimum is at π/2 and the maximum is I believe 7π/4
– Seungsoo Im
Nov 23 at 23:18
nevermind, the maximum value is when t = 3π/2
– Seungsoo Im
Nov 24 at 0:35
add a comment |
Here's a plot of your function (blue) and its derivative (orange):
Because the derivative is always positive, the minimum of your function is at the left (at $pi/4$) and the maximum is at the right (at $7 pi/4$).
$f(t) = 9 t + cot left( {t over 2} right)$
and
${partial over partial t} f(t) = 9 - {1 over 2} csc^2 left( {t over 2} right)$
answered Nov 20 at 7:03
David G. Stork
9,69621232
Here's a plot of your function (blue) and its derivative (orange):
Because the derivative is always positive, the minimum of your function is at the left (at $pi/4$) and the maximum is at the right (at $7 pi/4$).
$f(t) = 9 t + cot left( {t over 2} right)$
and
${partial over partial t} f(t) = 9 - {1 over 2} csc^2 left( {t over 2} right)$
answered Nov 20 at 7:03
David G. Stork
9,69621232
answered Nov 20 at 7:03
David G. Stork
9,69621232
answered Nov 20 at 7:03
David G. Stork
9,69621232
answered Nov 20 at 7:03
David G. Stork
9,69621232
9,69621232
wait.. according to my answers, the minimum is at π/2 and the maximum is I believe 7π/4
– Seungsoo Im
Nov 23 at 23:18
nevermind, the maximum value is when t = 3π/2
– Seungsoo Im
Nov 24 at 0:35
add a comment |
wait.. according to my answers, the minimum is at π/2 and the maximum is I believe 7π/4
– Seungsoo Im
Nov 23 at 23:18
nevermind, the maximum value is when t = 3π/2
– Seungsoo Im
Nov 24 at 0:35
wait.. according to my answers, the minimum is at π/2 and the maximum is I believe 7π/4
– Seungsoo Im
Nov 23 at 23:18
wait.. according to my answers, the minimum is at π/2 and the maximum is I believe 7π/4
– Seungsoo Im
Nov 23 at 23:18
nevermind, the maximum value is when t = 3π/2
– Seungsoo Im
Nov 24 at 0:35
nevermind, the maximum value is when t = 3π/2
– Seungsoo Im
Nov 24 at 0:35
add a comment |
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Check your derivative.
– David G. Stork
Nov 20 at 6:58
its right although it can definitely be simplified further.
– Seungsoo Im
Nov 23 at 23:16