Modelling Curves as-level

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A0W0N
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#1
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#1
I have this question which i have started to attempt but i don't think i'm right i'd appreciate some help. I'll pos tthe question along with my attempt
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A0W0N
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#2
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#2
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For i)
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I don't know where i was going with this
for ii)
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i guessing I'll need to differentiate it twice but then i don't onow what points to put into the equation
for iii) i have no idea how I'm supposed to draw it, i could do 8e^x or -4e^2x on its own but together i don't know

I'd really appreciate any help
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RDKGames
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#3
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#3
(Original post by A0W0N)
For i)

I don't know where i was going with this
for ii)

i guessing I'll need to differentiate it twice but then i don't onow what points to put into the equation
for iii) i have no idea how I'm supposed to draw it, i could do 8e^x or -4e^2x on its own but together i don't know

I'd really appreciate any help
So, which log rule are you using when going from

8e^x - 4e^{2x} = 0

to

8\ln (x) - 4 \ln(2x) = 0

??
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A0W0N
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#4
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#4
(Original post by RDKGames)
So, which log rule are you using when going from

8e^x - 4e^{2x} = 0

to

8\ln (x) - 4 \ln(2x) = 0

??
ok i understand where i went wrong, i can't put ln(0) so wgat am i supposed to do to solve this?
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RDKGames
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#5
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#5
(Original post by A0W0N)
ok i understand where i went wrong, i can't put ln(0) so wgat am i supposed to do to solve this?
Maybe move one of the exponentials over to the RHS so you don't end up taking log of 0.

But more generally, don't be under the impression that \log(a+b) = \log(a) + \log (b).
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Sophhhowa
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#6
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For the first part it also says axes so you also have to consider when x=0
ii) set 1st derivative = 0, then show that x=0
Are you Ok with the sketch?
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A0W0N
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#7
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#7
(Original post by Sophhhowa)
For the first part it also says axes so you also have to consider when x=0
ii) set 1st derivative = 0, then show that x=0
Are you Ok with the sketch?
I have no clue what I'm supposed to do for the sketch
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A0W0N
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#8
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#8
(Original post by RDKGames)
Maybe move one of the exponentials over to the RHS so you don't end up taking log of 0.

But more generally, don't be under the impression that \log(a+b) = \log(a) + \log (b).
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i got this again
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RDKGames
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#9
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#9
(Original post by A0W0N)
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i got this again
ln(...) is an operation you apply to both sides of the equation.

Point being is that you must apply it to BOTH sides of the equation ENTIRELY.

So from

4e^(2x) = 8e^x

just divide both sides by 4 first to get

e^(2x) = 2e^x

then log BOTH SIDES ENTIRELY to obtain

ln( e^(2x) ) = ln( 2e^x )

and NOT

ln( e^(2x) ) = 2 ln( e^x )

which is a big error you are making.
Last edited by RDKGames; 2 years ago
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A0W0N
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#10
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ok i now understand what you meant and the mistake i made can you help me with the other two part please?
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RDKGames
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#11
(Original post by A0W0N)
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ok i now understand what you meant and the mistake i made can you help me with the other two part please?
Looks good, just remember my comment about getting out of habit writing decimals...

For part (ii), well a good starting point is asking yourself if you know what a turning point is first of all.
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A0W0N
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#12
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#12
(Original post by RDKGames)
Looks good, just remember my comment about getting out of habit writing decimals...

For part (ii), well a good starting point is asking yourself if you know what a turning point is first of all.
Isn't a turning point where the lin changes direction?
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RDKGames
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#13
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#13
(Original post by A0W0N)
Isn't a turning point where the lin changes direction?
You can be more specific than that.

It's where the curve goes from increasing to decreasing, or vice versa. Ultimately, it is the point where the first derivative is zero, and if the second derivative is non-zero then it is a turning point (as opposed to a point of inflection)
Last edited by RDKGames; 2 years ago
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A0W0N
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#14
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I've somehow messed this up
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Sophhhowa
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#15
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#15
(Original post by A0W0N)
I have no clue what I'm supposed to do for the sketch
Think about the key points. E.g turning points, where the graph meets any axes and the general trend
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RDKGames
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#16
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#16
(Original post by A0W0N)
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I've somehow messed this up
No, you haven't.

What is the one value of x which makes this equation hold?
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A0W0N
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#17
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#17
(Original post by RDKGames)
No, you haven't.

What is the one value of x which makes this equation hold?
So the turning point is at (0.0)?
Last edited by A0W0N; 2 years ago
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RDKGames
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#18
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#18
(Original post by A0W0N)
So the turning point is at (0.0)?
x=0 is correct but you should check that y coordinate because it's not zero.

Remember that e^0 is equal to 1.
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A0W0N
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#19
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#19
i have the points is it still like a ln graph? i'm guessing its going to be like this, but I'm not sure
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RDKGames
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#20
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#20
(Original post by A0W0N)
i have the points is it still like a ln graph? i'm guessing its going to be like this, but I'm not sure
Yes you got the correct sketch now.
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