ihatemaths000
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#1
Report Thread starter 11 years ago
#1
what's the range of g(x)= x^2 - 4x + 11


f(x) = |x-2|-3

what is gf(x)
what is gf(-1)

:tsr2:
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Sir Cumference
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#2
Report 11 years ago
#2
For the first one: complete the square.
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*BCM*
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#3
Report 11 years ago
#3
you havent specified the domain (what values x can take, so range is impossible to define)

for 2nd part, gf(x)=(|x-2|-3)^2 -4(|x-2|-3) + 11 and solve

ull get 2 possible sets of reults for x<2 and for x>2
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dezlee
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#4
Report 11 years ago
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Lol, for the first one:

range is values in which f(x) can take.
so if you imagine a x^2 graph, what vaues of y can be taken?

For the second:

for gf(-1)................. sub -1 into requation of f(x) and sub f(x) into g(x)?
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*BCM*
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#5
Report 11 years ago
#5
(Original post by ihatemaths000)
so gf(-1) = (|-3|-3)^2 - 4(|-3|-3) + 11

but i dont know what to do with the modulus bits when the brackets are times out
modulus just make everything inside positive, so |-3|=3
so

gf(-1)=0^2 -4*0 +11=11

RE: range: for x>_0, y>_(min value y can take)

to find the minimum value you either differentiate the function or complete the square:

EITHER:
g(x)=x^2 - 4x + 11
g'(x)=2x - 4=0 (youre finding min point so differential=0)
x=2
y is min at x=2
so ymin=g(2)=4 - 8 + 11=7

so range is y>_7

OR
g(x)=x^2 - 4x + 11=
=x^2 - 4x + 4 + 7=
=(x - 2)^2 + 7

g(x)=min when x=2
min value g(x) can take is 7

so range is y>_7



EDIT: ">_" = greater than or equals to
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