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"here are two statements about a stationary value for the function f(x)=4sinx-2;
1) f has a stationary value when x= (π/3)
2) f has a stationary value when x=(π/2)
I simplified the equation so that it was equal to 1/2 (one half).
I then looked at the exact value of sin 1/2; which is equal to 30 degrees.
I thought that meant that neither statement was correct; because 1) is equal to 60 degrees; and 2) is equal to 90 degrees.
However, the answers specify that 2) is correct.
1) f has a stationary value when x= (π/3)
2) f has a stationary value when x=(π/2)
I simplified the equation so that it was equal to 1/2 (one half).
I then looked at the exact value of sin 1/2; which is equal to 30 degrees.
I thought that meant that neither statement was correct; because 1) is equal to 60 degrees; and 2) is equal to 90 degrees.
However, the answers specify that 2) is correct.
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#3
(Original post by apronedsamurai)
I simplified the equation so that it was equal to 1/2 (one half).
I then looked at the exact value of sin 1/2; which is equal to 30 degrees.
I simplified the equation so that it was equal to 1/2 (one half).
I then looked at the exact value of sin 1/2; which is equal to 30 degrees.
What would you normally do to find the stationary points of a curve?
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#4
If this is A2, you need to differentiate the function and find where the differential is equal to 0.
If this is AS, don't worry, look and the sine graph and find the points where the graph is parallel to the x axis
If this is AS, don't worry, look and the sine graph and find the points where the graph is parallel to the x axis
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I thought 4sinx-2=0
4sinx=2
sinx=2/4 simplified to 1/2
sin 1/2 (30 degrees)
Err, differentiate? I thought that might be the case, but that I had no idea how to do that....4sin?
Or would it differentiate to simply sin; meaning that the normal sin function is used, where 4 would be the maximum point; and so it would pass through the 90 degree mark?
4sinx=2
sinx=2/4 simplified to 1/2
sin 1/2 (30 degrees)
Err, differentiate? I thought that might be the case, but that I had no idea how to do that....4sin?
Or would it differentiate to simply sin; meaning that the normal sin function is used, where 4 would be the maximum point; and so it would pass through the 90 degree mark?
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#8
(Original post by apronedsamurai)
No graph is provided.
No graph is provided.
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#9
(Original post by apronedsamurai)
I thought 4sinx-2=0
4sinx=2
sinx=2/4 simplified to 1/2
sin 1/2 (30 degrees)
Err, differentiate? I thought that might be the case, but that I had no idea how to do that....4sin?
I thought 4sinx-2=0
4sinx=2
sinx=2/4 simplified to 1/2
sin 1/2 (30 degrees)
Err, differentiate? I thought that might be the case, but that I had no idea how to do that....4sin?
Do you know what the graph of Sin(x) looks like?
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Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
Yes, I know what the (original, i.e. unmodified) graph of Sin(x) looks like.
Maximum value of 1; Min. value of -1; intersects at 90 and 180 degrees respectively.
This is "Higher"
Yes, I know what the (original, i.e. unmodified) graph of Sin(x) looks like.
Maximum value of 1; Min. value of -1; intersects at 90 and 180 degrees respectively.
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#11
(Original post by apronedsamurai)
Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
Can you differentiate Sin(x)
If Not
Do you know what the graph of Sin(x) looks like?
Do you know what a stationary point is?
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#12
(Original post by apronedsamurai)
Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
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#13
(Original post by apronedsamurai)
Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
Yes, I know what the (original, i.e. unmodified) graph of Sin(x) looks like.
Maximum value of 1; Min. value of -1; intersects at 90 and 180 degrees respectively.
Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
Yes, I know what the (original, i.e. unmodified) graph of Sin(x) looks like.
Maximum value of 1; Min. value of -1; intersects at 90 and 180 degrees respectively.
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#14
So sketch a graph of the translated sin(x) graph and use it to find the maximum and minimum points.
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#15
(Original post by apronedsamurai)
Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
Yes, I know what the (original, i.e. unmodified) graph of Sin(x) looks like.
Maximum value of 1; Min. value of -1; intersects at 90 and 180 degrees respectively.
Im in Scotland, so dont know what AS and AS2 are.
This is "Higher"
Yes, I know what the (original, i.e. unmodified) graph of Sin(x) looks like.
Maximum value of 1; Min. value of -1; intersects at 90 and 180 degrees respectively.
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#16
(Original post by davros)
What definition does your textbook use for "stationary point"? You're almost certainly expected to use differentiation for this!
What definition does your textbook use for "stationary point"? You're almost certainly expected to use differentiation for this!
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#17
(Original post by morgan8002)
I don't think they know how to differentiate trig functions. Either they haven't covered it yet and they are trying to do questions that they haven't covered the methods for, or the question has to be solved graphically.
I don't think they know how to differentiate trig functions. Either they haven't covered it yet and they are trying to do questions that they haven't covered the methods for, or the question has to be solved graphically.
There's no point in introducing the concept of "stationary point" prior to covering differentiation, because there isn't going to be a systematic way of tackling questions like this without that (obviously you "get lucky" with trig functions because the derivative of sin is just cos, which you know how to inspect!).
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Can this thread be closed please? Im getting a bit uncomfortable with it.
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#19
(Original post by davros)
That's what I'm trying to get the OP to communicate.
There's no point in introducing the concept of "stationary point" prior to covering differentiation, because there isn't going to be a systematic way of tackling questions like this without that (obviously you "get lucky" with trig functions because the derivative of sin is just cos, which you know how to inspect!).
That's what I'm trying to get the OP to communicate.
There's no point in introducing the concept of "stationary point" prior to covering differentiation, because there isn't going to be a systematic way of tackling questions like this without that (obviously you "get lucky" with trig functions because the derivative of sin is just cos, which you know how to inspect!).
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#20
(Original post by apronedsamurai)
Can this thread be closed please? Im getting a bit uncomfortable with it.
Can this thread be closed please? Im getting a bit uncomfortable with it.
If you just tell us what your knowledge of stationary points is we can help
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