Modelling Curves as-level

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 :smile:

Reply 2

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 :smile:

So, which log rule are you using when going from

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

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

Reply 3

Original post by RDKGames

So, which log rule are you using when going from

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

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?

Reply 4

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)

Reply 5

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?

Reply 6

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

Reply 7

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)

i got this again

Reply 8

Original post by A0W0N

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.

(edited 3 years ago)

Reply 9

ok i now understand what you meant and the mistake i made can you help me with the other two part please?

Reply 10

Original post by A0W0N

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.

Reply 11

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?

Reply 12