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The set of values of $a$ for which the function does not posses critical points
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The set of all values of '$a$' for which the function,
$f(x)=(a^2-3a+2)(cos^2x/4 - sin^2x/4) + (a-1)x + sin1$ does not posses critical points is:
I first differented it to find $f'(x)$, then I tried to find out the condition where it becomes zero so that the answer will be $mathrmR -$ (the range). This is because for critical points, the function shouldn't be zero or not defined or non differentiable. It seems differentiable and could not have infinite slope.
But after differenting, I got:
$(a^2 - 3a +2)(-dfrac sinx/22) + (a-1)$
I don't get how to proceed with two variables $x$ and $a$.
maxima-minima
asked Jul 15 at 13:10
Ice Inkberry
203111
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up vote
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The set of all values of '$a$' for which the function,
$f(x)=(a^2-3a+2)(cos^2x/4 - sin^2x/4) + (a-1)x + sin1$ does not posses critical points is:
I first differented it to find $f'(x)$, then I tried to find out the condition where it becomes zero so that the answer will be $mathrmR -$ (the range). This is because for critical points, the function shouldn't be zero or not defined or non differentiable. It seems differentiable and could not have infinite slope.
But after differenting, I got:
$(a^2 - 3a +2)(-dfrac sinx/22) + (a-1)$
I don't get how to proceed with two variables $x$ and $a$.
maxima-minima
asked Jul 15 at 13:10
Ice Inkberry
203111
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The set of all values of '$a$' for which the function,
$f(x)=(a^2-3a+2)(cos^2x/4 - sin^2x/4) + (a-1)x + sin1$ does not posses critical points is:
I first differented it to find $f'(x)$, then I tried to find out the condition where it becomes zero so that the answer will be $mathrmR -$ (the range). This is because for critical points, the function shouldn't be zero or not defined or non differentiable. It seems differentiable and could not have infinite slope.
But after differenting, I got:
$(a^2 - 3a +2)(-dfrac sinx/22) + (a-1)$
I don't get how to proceed with two variables $x$ and $a$.
maxima-minima
asked Jul 15 at 13:10
Ice Inkberry
203111
The set of all values of '$a$' for which the function,
$f(x)=(a^2-3a+2)(cos^2x/4 - sin^2x/4) + (a-1)x + sin1$ does not posses critical points is:
I first differented it to find $f'(x)$, then I tried to find out the condition where it becomes zero so that the answer will be $mathrmR -$ (the range). This is because for critical points, the function shouldn't be zero or not defined or non differentiable. It seems differentiable and could not have infinite slope.
But after differenting, I got:
$(a^2 - 3a +2)(-dfrac sinx/22) + (a-1)$
I don't get how to proceed with two variables $x$ and $a$.
maxima-minima
asked Jul 15 at 13:10
Ice Inkberry
203111
edited Jul 15 at 13:23
asked Jul 15 at 13:10
Ice Inkberry
203111
asked Jul 15 at 13:10
Ice Inkberry
203111
asked Jul 15 at 13:10
Ice Inkberry
203111
203111
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
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Check your first derivative i have got
$$(a^2-3a+2)left(frac-sin(2x)2right)+a-1$$ and then write
$$(a-1)(a-2)sin(2x)=2(a-1)$$
Can you finish?
answered Jul 15 at 13:20
Dr. Sonnhard Graubner
66.9k32659
1
and i added a hint for solving the Problem!
– Dr. Sonnhard Graubner
Jul 15 at 13:23
add a comment |Â
up vote
1
down vote
Notice that if $a=1$, then $f$ is a constant function.
Now, we focus on $a ne 1$,
$$(a^2-3a+2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-1)(a-2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-2) left( - fracsin x/ 22right)+1=0$$
If $a=2$, then the equation has no solution.
Otherwise, $$sin x/2 =- frac2a-2$$
It doesn't have a solution if if $$frac2 > 1$$
Hopefully you can carry on from here.
answered Jul 15 at 13:17
Siong Thye Goh
77.8k134796
I did notice about $a =1$ case but couldn't think of more. Thankyou!
– Ice Inkberry
Jul 15 at 13:20
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Check your first derivative i have got
$$(a^2-3a+2)left(frac-sin(2x)2right)+a-1$$ and then write
$$(a-1)(a-2)sin(2x)=2(a-1)$$
Can you finish?
answered Jul 15 at 13:20
Dr. Sonnhard Graubner
66.9k32659
1
and i added a hint for solving the Problem!
– Dr. Sonnhard Graubner
Jul 15 at 13:23
add a comment |Â
up vote
1
down vote
Check your first derivative i have got
$$(a^2-3a+2)left(frac-sin(2x)2right)+a-1$$ and then write
$$(a-1)(a-2)sin(2x)=2(a-1)$$
Can you finish?
answered Jul 15 at 13:20
Dr. Sonnhard Graubner
66.9k32659
1
and i added a hint for solving the Problem!
– Dr. Sonnhard Graubner
Jul 15 at 13:23
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Check your first derivative i have got
$$(a^2-3a+2)left(frac-sin(2x)2right)+a-1$$ and then write
$$(a-1)(a-2)sin(2x)=2(a-1)$$
Can you finish?
answered Jul 15 at 13:20
Dr. Sonnhard Graubner
66.9k32659
Check your first derivative i have got
$$(a^2-3a+2)left(frac-sin(2x)2right)+a-1$$ and then write
$$(a-1)(a-2)sin(2x)=2(a-1)$$
Can you finish?
answered Jul 15 at 13:20
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 13:20
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 13:20
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 13:20
Dr. Sonnhard Graubner
66.9k32659
66.9k32659
1
and i added a hint for solving the Problem!
– Dr. Sonnhard Graubner
Jul 15 at 13:23
add a comment |Â
1
and i added a hint for solving the Problem!
– Dr. Sonnhard Graubner
Jul 15 at 13:23
1
1
and i added a hint for solving the Problem!
– Dr. Sonnhard Graubner
Jul 15 at 13:23
and i added a hint for solving the Problem!
– Dr. Sonnhard Graubner
Jul 15 at 13:23
add a comment |Â
up vote
1
down vote
Notice that if $a=1$, then $f$ is a constant function.
Now, we focus on $a ne 1$,
$$(a^2-3a+2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-1)(a-2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-2) left( - fracsin x/ 22right)+1=0$$
If $a=2$, then the equation has no solution.
Otherwise, $$sin x/2 =- frac2a-2$$
It doesn't have a solution if if $$frac2 > 1$$
Hopefully you can carry on from here.
answered Jul 15 at 13:17
Siong Thye Goh
77.8k134796
I did notice about $a =1$ case but couldn't think of more. Thankyou!
– Ice Inkberry
Jul 15 at 13:20
add a comment |Â
up vote
1
down vote
Notice that if $a=1$, then $f$ is a constant function.
Now, we focus on $a ne 1$,
$$(a^2-3a+2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-1)(a-2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-2) left( - fracsin x/ 22right)+1=0$$
If $a=2$, then the equation has no solution.
Otherwise, $$sin x/2 =- frac2a-2$$
It doesn't have a solution if if $$frac2 > 1$$
Hopefully you can carry on from here.
answered Jul 15 at 13:17
Siong Thye Goh
77.8k134796
I did notice about $a =1$ case but couldn't think of more. Thankyou!
– Ice Inkberry
Jul 15 at 13:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Notice that if $a=1$, then $f$ is a constant function.
Now, we focus on $a ne 1$,
$$(a^2-3a+2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-1)(a-2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-2) left( - fracsin x/ 22right)+1=0$$
If $a=2$, then the equation has no solution.
Otherwise, $$sin x/2 =- frac2a-2$$
It doesn't have a solution if if $$frac2 > 1$$
Hopefully you can carry on from here.
answered Jul 15 at 13:17
Siong Thye Goh
77.8k134796
Notice that if $a=1$, then $f$ is a constant function.
Now, we focus on $a ne 1$,
$$(a^2-3a+2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-1)(a-2)left(- fracsin x/ 22right)+(a-1)=0$$
$$(a-2) left( - fracsin x/ 22right)+1=0$$
If $a=2$, then the equation has no solution.
Otherwise, $$sin x/2 =- frac2a-2$$
It doesn't have a solution if if $$frac2 > 1$$
Hopefully you can carry on from here.
answered Jul 15 at 13:17
Siong Thye Goh
77.8k134796
edited Jul 15 at 13:28
answered Jul 15 at 13:17
Siong Thye Goh
77.8k134796
answered Jul 15 at 13:17
Siong Thye Goh
77.8k134796
answered Jul 15 at 13:17
Siong Thye Goh
77.8k134796
77.8k134796
I did notice about $a =1$ case but couldn't think of more. Thankyou!
– Ice Inkberry
Jul 15 at 13:20
add a comment |Â
I did notice about $a =1$ case but couldn't think of more. Thankyou!
– Ice Inkberry
Jul 15 at 13:20
I did notice about $a =1$ case but couldn't think of more. Thankyou!
– Ice Inkberry
Jul 15 at 13:20
I did notice about $a =1$ case but couldn't think of more. Thankyou!
– Ice Inkberry
Jul 15 at 13:20
add a comment |Â
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