- #1
dylanjames
- 24
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New to composite functions here. Lesson has been vague and unhelpful.. again. Here is what I've worked on so far but not sure on the last equation in particular, or that I have done my multiplication properly when working with a squared set of brackets, multiplied by an number.. (b and c)
Any confirmation/help would be mucho appreciated. Thanks!
Given the functions f(x)=4x-1, g(x)=3-2^2 and h(x)=√(x+5) simplify the following..
a) f(g(a))
g(a)=4(3-2a^2) - 1
=12 - 8a^2 - 1
=11-8a^2.
b) g(f(2x))
f(2x)=4(2x)-1=8x-1
g(f(2x)=3-2(8x-1)^2
g(f(2x)=3-2[(8x-1)(8x-1)]
g(f(2x))=3-2[64x^2 -16x+1]
g(f(2x)=-128x^2+32x-2+3
g(f(2x)=-128x^2+32x+1
c) h(g(2k+1))
g(2k+1)=3-2(2k+1)^2
=3-2[(2k+1)(2k+1)]
=3-2[4k^2 +4k+1]
=-8k^2-8k+1
h(g(2k+1))=sqrt[(-8k^2-8k+1)+5] =?
=-8k-2.82k+2.23? = 10.82k+2.23...
That can't be right...
Also need the domain and range for f(g(x))... no idea how to tackle that.
Any confirmation/help would be mucho appreciated. Thanks!
Given the functions f(x)=4x-1, g(x)=3-2^2 and h(x)=√(x+5) simplify the following..
a) f(g(a))
g(a)=4(3-2a^2) - 1
=12 - 8a^2 - 1
=11-8a^2.
b) g(f(2x))
f(2x)=4(2x)-1=8x-1
g(f(2x)=3-2(8x-1)^2
g(f(2x)=3-2[(8x-1)(8x-1)]
g(f(2x))=3-2[64x^2 -16x+1]
g(f(2x)=-128x^2+32x-2+3
g(f(2x)=-128x^2+32x+1
c) h(g(2k+1))
g(2k+1)=3-2(2k+1)^2
=3-2[(2k+1)(2k+1)]
=3-2[4k^2 +4k+1]
=-8k^2-8k+1
h(g(2k+1))=sqrt[(-8k^2-8k+1)+5] =?
=-8k-2.82k+2.23? = 10.82k+2.23...
That can't be right...
Also need the domain and range for f(g(x))... no idea how to tackle that.