#A level maths help
61 messages · Page 1 of 1 (latest)
so just to like point out
obviously the quadratic in a is the denominator in b so g(x) you can write as
1/(3(x-2)^2 + 8)
right
yep
now that (x-2)^2
what can you tell me about the range of that
what can you tell me about the output of it
that it has to be positive
yep or 0
now 3 times that obviously wont make a difference
ah not quite so
ok
lowest actually isnt right i was wrong
thats fine
so if you made the 3(x-2)^2 positive
say instead it was like 3
your g(x) would be 1/11 right
so smaller than your 1/8
say instead what was 3?
your 3(x-2)^2
oh ok yhyh
just pretend you put in a value so it was 3
ok
then your results gonna be 1/11
smaller than your 1/8
what if you plugged in a larger number
like as you make 3(x-2)^2 bigger and bigger
whats gonna happen to your 1/(3(x-2)^2 + 8)
that itll just get smaller and smaller
0?
ah ok
itll approach 0
so is the lim thing?
so g(x) is at most 1/8
ok
but also greater than 0
perfect
no worries!