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Finding the minimum of a
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For constant number "a" consider the function $f(x)= ax + cos 2x + sin x + cos x$ on $mathbb R$ such that $f(u)<f(v)$ for $u< v$. If the range of a, for any real numbers $u, v$ is $[dfracmn, infty)$. Find the minimum value of $(m+n)$
Attempt:
The problem is just an overly complicated way of saying "find the range for which f(x) is strictly increasing".
So, using $f'(x)>0$
I get:
$a> 2sin 2x + sin x - cos x$
So basically we need to find the maximum of RHS for the range of a.
$g(x)= 2 sin 2x + sin x - cos x $
For extremum, $g'(x)=0 $
$4 cos 2x + cos x+ sin x= 0$
$implies 16cos^2 2x = 1+ sin 2x$
$implies 16sin^2 2x + sin 2x -15 =0$
$implies sin 2x = -1$ or $sin 2x = dfrac 1516$
Then it seriously gets very complicated because we would have to extract $sin x$ and $cos x$ from $sin 2x = dfrac1516$. I tried but couldn't get it easily. After this we would have to plug in the values to g(x)
to see which one of them gives maximum. (Just a side note: g is periodic with period $2pi$)
Is it possible to do this problem using my method? If not, what are the other clever ways to solve it?
calculus trigonometry
asked Jul 15 at 16:32
Abcd
2,3831624
add a comment |Â
up vote
4
down vote
favorite
For constant number "a" consider the function $f(x)= ax + cos 2x + sin x + cos x$ on $mathbb R$ such that $f(u)<f(v)$ for $u< v$. If the range of a, for any real numbers $u, v$ is $[dfracmn, infty)$. Find the minimum value of $(m+n)$
Attempt:
The problem is just an overly complicated way of saying "find the range for which f(x) is strictly increasing".
So, using $f'(x)>0$
I get:
$a> 2sin 2x + sin x - cos x$
So basically we need to find the maximum of RHS for the range of a.
$g(x)= 2 sin 2x + sin x - cos x $
For extremum, $g'(x)=0 $
$4 cos 2x + cos x+ sin x= 0$
$implies 16cos^2 2x = 1+ sin 2x$
$implies 16sin^2 2x + sin 2x -15 =0$
$implies sin 2x = -1$ or $sin 2x = dfrac 1516$
Then it seriously gets very complicated because we would have to extract $sin x$ and $cos x$ from $sin 2x = dfrac1516$. I tried but couldn't get it easily. After this we would have to plug in the values to g(x)
to see which one of them gives maximum. (Just a side note: g is periodic with period $2pi$)
Is it possible to do this problem using my method? If not, what are the other clever ways to solve it?
calculus trigonometry
asked Jul 15 at 16:32
Abcd
2,3831624
You've done very well so far, but I think you've lost track of what you are trying to accomplish. You want to find the maximum value of $g(x)$. Can you find that if you know the associated value of $sin 2x?$
– saulspatz
Jul 15 at 17:16
Notice that there is no minimum value for $m + n$ unless we require $m, n$ to be positive integers.
– Luca Bressan
Jul 15 at 17:31
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
For constant number "a" consider the function $f(x)= ax + cos 2x + sin x + cos x$ on $mathbb R$ such that $f(u)<f(v)$ for $u< v$. If the range of a, for any real numbers $u, v$ is $[dfracmn, infty)$. Find the minimum value of $(m+n)$
Attempt:
The problem is just an overly complicated way of saying "find the range for which f(x) is strictly increasing".
So, using $f'(x)>0$
I get:
$a> 2sin 2x + sin x - cos x$
So basically we need to find the maximum of RHS for the range of a.
$g(x)= 2 sin 2x + sin x - cos x $
For extremum, $g'(x)=0 $
$4 cos 2x + cos x+ sin x= 0$
$implies 16cos^2 2x = 1+ sin 2x$
$implies 16sin^2 2x + sin 2x -15 =0$
$implies sin 2x = -1$ or $sin 2x = dfrac 1516$
Then it seriously gets very complicated because we would have to extract $sin x$ and $cos x$ from $sin 2x = dfrac1516$. I tried but couldn't get it easily. After this we would have to plug in the values to g(x)
to see which one of them gives maximum. (Just a side note: g is periodic with period $2pi$)
Is it possible to do this problem using my method? If not, what are the other clever ways to solve it?
calculus trigonometry
asked Jul 15 at 16:32
Abcd
2,3831624
For constant number "a" consider the function $f(x)= ax + cos 2x + sin x + cos x$ on $mathbb R$ such that $f(u)<f(v)$ for $u< v$. If the range of a, for any real numbers $u, v$ is $[dfracmn, infty)$. Find the minimum value of $(m+n)$
Attempt:
The problem is just an overly complicated way of saying "find the range for which f(x) is strictly increasing".
So, using $f'(x)>0$
I get:
$a> 2sin 2x + sin x - cos x$
So basically we need to find the maximum of RHS for the range of a.
$g(x)= 2 sin 2x + sin x - cos x $
For extremum, $g'(x)=0 $
$4 cos 2x + cos x+ sin x= 0$
$implies 16cos^2 2x = 1+ sin 2x$
$implies 16sin^2 2x + sin 2x -15 =0$
$implies sin 2x = -1$ or $sin 2x = dfrac 1516$
Then it seriously gets very complicated because we would have to extract $sin x$ and $cos x$ from $sin 2x = dfrac1516$. I tried but couldn't get it easily. After this we would have to plug in the values to g(x)
to see which one of them gives maximum. (Just a side note: g is periodic with period $2pi$)
Is it possible to do this problem using my method? If not, what are the other clever ways to solve it?
calculus trigonometry
asked Jul 15 at 16:32
Abcd
2,3831624
edited Jul 15 at 17:02
asked Jul 15 at 16:32
Abcd
2,3831624
asked Jul 15 at 16:32
Abcd
2,3831624
asked Jul 15 at 16:32
Abcd
2,3831624
2,3831624
You've done very well so far, but I think you've lost track of what you are trying to accomplish. You want to find the maximum value of $g(x)$. Can you find that if you know the associated value of $sin 2x?$
– saulspatz
Jul 15 at 17:16
Notice that there is no minimum value for $m + n$ unless we require $m, n$ to be positive integers.
– Luca Bressan
Jul 15 at 17:31
add a comment |Â
You've done very well so far, but I think you've lost track of what you are trying to accomplish. You want to find the maximum value of $g(x)$. Can you find that if you know the associated value of $sin 2x?$
– saulspatz
Jul 15 at 17:16
Notice that there is no minimum value for $m + n$ unless we require $m, n$ to be positive integers.
– Luca Bressan
Jul 15 at 17:31
You've done very well so far, but I think you've lost track of what you are trying to accomplish. You want to find the maximum value of $g(x)$. Can you find that if you know the associated value of $sin 2x?$
– saulspatz
Jul 15 at 17:16
You've done very well so far, but I think you've lost track of what you are trying to accomplish. You want to find the maximum value of $g(x)$. Can you find that if you know the associated value of $sin 2x?$
– saulspatz
Jul 15 at 17:16
Notice that there is no minimum value for $m + n$ unless we require $m, n$ to be positive integers.
– Luca Bressan
Jul 15 at 17:31
Notice that there is no minimum value for $m + n$ unless we require $m, n$ to be positive integers.
– Luca Bressan
Jul 15 at 17:31
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
You want to maximize $$a=g(x)= 2sin 2x + sin x - cos x=2sin 2x + sqrt1-sin 2x$$ (I'm taking the plus sign on the square root because we only care about the maximum.) You've found the possible values of $sin 2x$ that maximize $a,$ so it should be plain sailing from here.
answered Jul 15 at 17:27
saulspatz
10.7k21323
Thanks.I was struggling because I just din't convert sin x- cos x to $sqrt 1-sin 2x$ :( .
– Abcd
Jul 15 at 17:38
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You want to maximize $$a=g(x)= 2sin 2x + sin x - cos x=2sin 2x + sqrt1-sin 2x$$ (I'm taking the plus sign on the square root because we only care about the maximum.) You've found the possible values of $sin 2x$ that maximize $a,$ so it should be plain sailing from here.
answered Jul 15 at 17:27
saulspatz
10.7k21323
Thanks.I was struggling because I just din't convert sin x- cos x to $sqrt 1-sin 2x$ :( .
– Abcd
Jul 15 at 17:38
add a comment |Â
up vote
3
down vote
accepted
You want to maximize $$a=g(x)= 2sin 2x + sin x - cos x=2sin 2x + sqrt1-sin 2x$$ (I'm taking the plus sign on the square root because we only care about the maximum.) You've found the possible values of $sin 2x$ that maximize $a,$ so it should be plain sailing from here.
answered Jul 15 at 17:27
saulspatz
10.7k21323
Thanks.I was struggling because I just din't convert sin x- cos x to $sqrt 1-sin 2x$ :( .
– Abcd
Jul 15 at 17:38
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You want to maximize $$a=g(x)= 2sin 2x + sin x - cos x=2sin 2x + sqrt1-sin 2x$$ (I'm taking the plus sign on the square root because we only care about the maximum.) You've found the possible values of $sin 2x$ that maximize $a,$ so it should be plain sailing from here.
answered Jul 15 at 17:27
saulspatz
10.7k21323
You want to maximize $$a=g(x)= 2sin 2x + sin x - cos x=2sin 2x + sqrt1-sin 2x$$ (I'm taking the plus sign on the square root because we only care about the maximum.) You've found the possible values of $sin 2x$ that maximize $a,$ so it should be plain sailing from here.
answered Jul 15 at 17:27
saulspatz
10.7k21323
answered Jul 15 at 17:27
saulspatz
10.7k21323
answered Jul 15 at 17:27
saulspatz
10.7k21323
answered Jul 15 at 17:27
saulspatz
10.7k21323
10.7k21323
Thanks.I was struggling because I just din't convert sin x- cos x to $sqrt 1-sin 2x$ :( .
– Abcd
Jul 15 at 17:38
add a comment |Â
Thanks.I was struggling because I just din't convert sin x- cos x to $sqrt 1-sin 2x$ :( .
– Abcd
Jul 15 at 17:38
Thanks.I was struggling because I just din't convert sin x- cos x to $sqrt 1-sin 2x$ :( .
– Abcd
Jul 15 at 17:38
Thanks.I was struggling because I just din't convert sin x- cos x to $sqrt 1-sin 2x$ :( .
– Abcd
Jul 15 at 17:38
add a comment |Â
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You've done very well so far, but I think you've lost track of what you are trying to accomplish. You want to find the maximum value of $g(x)$. Can you find that if you know the associated value of $sin 2x?$
– saulspatz
Jul 15 at 17:16
Notice that there is no minimum value for $m + n$ unless we require $m, n$ to be positive integers.
– Luca Bressan
Jul 15 at 17:31