3U Maths Question! (1 Viewer)

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its easy ( just long )

go through the usual "menu"

find intercepts, find the domain , differentiate and set to zero to find turning points, classify turning points, then get second derivative and find inflexions ( remember to check a change in concavity about possible inflexion points ) , then sub in a few values of x to get a few more points on the graph, then draw
 
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or you try doing ( 3 +e^(x) - 3 ) / (3 +e^(x) )
= 1 - [3 / ( 3+e^(x) ]

might be a bit easier, just graph the second part and shift it all up by 1

but thats bit dodgy
 

muzeikchun852

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its easy ( just long )

go through the usual "menu"

find intercepts, find the domain , differentiate and set to zero to find turning points, classify turning points, then get second derivative and find inflexions ( remember to check a change in concavity about possible inflexion points ) , then sub in a few values of x to get a few more points on the graph, then draw
i derived it and i got:

which doesnt help.

there isnt any asymptote so i can use the 4U method.
 
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i derived it and i got:

which doesnt help.
no, that just means theres no turning points, I just check wolfram and it shows that it does have a soln for d^2 y / dx^2 =0

meaning it could have a possible inflexion point.

It you dont get anything you just go onto the next step.
 
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take the limits of f(x) as x goes to positive infinity and as x goes to negative infinity

as x-> + infinity , f(x) -> 1

as x-> - infinity , f(x) -> 0

these are asymtotes

also you will probably need to pick some random of values of x, and just sub then in to a general shape of the graph
 
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lol u dont need 4unit graphing for this, its jsut a slightly harder 3unit graph where u will need to sub a few values to x to see whats happening
 

muzeikchun852

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take the limits of f(x) as x goes to positive infinity and as x goes to negative infinity

as x-> + infinity , f(x) -> 1

as x-> - infinity , f(x) -> 0

these are asymtotes

also you will probably need to pick some random of values of x, and just sub then in to a general shape of the graph
k. thankyou :]
 

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