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December 21, 2014

December 21, 2014

Posted by **Jess** on Tuesday, April 28, 2009 at 7:11am.

f(x) = -3x + 9

g(x) = 8x + 7

Find g(f(x))

I got -24x + 16.

Another :

f(x) = -3x + 9

g(x) = 8x + 7

Find f(g(x))

Thanks in advance.

- Math - Algebra II -
**Reiny**, Tuesday, April 28, 2009 at 7:50amno, one term was right, but your constant is wrong.

do it this way :

f(x) = -3x + 9

g(x) = 8x + 7

then g(f(x))

= g(-3x+9)

= 8(-3x+9) + 7

= -24x + 72 + 7

= -24x + 79

and f(g(x))

= f(8x+7)

= -3(8x+7) + 9

= -24x - 21 + 9

= -24x - 12

- Math - Algebra II -
**Jess**, Tuesday, April 28, 2009 at 8:15amThank you so much! I've been sitting here for hours. Only 4 to go. Thanks :)

- Math - Algebra II -
**Jess**, Tuesday, April 28, 2009 at 8:40am2 down.

I'm having trouble here

f(x) = 4x + 2

h(x) = -7x - 5

Find f(h(x))

I tried doing it how it was done above, but i'm not getting an answer listed.

= f(-7x - 5)

= 4x(-7x-5)+ 2

= 28x^2-20+2

= 28x^2-18

Where am I going wrong :| ?

ALSO -

f(x) = 4x-3

g(x) = x2-2

Find f(3a-4)

- Math - Algebra II -
**Reiny**, Tuesday, April 28, 2009 at 8:58amfrom

= f(-7x - 5) to

= 4x(-7x-5)+ 2

you are replacing the x, so don't leave the x there.

should be

= 4(-7x-5)+ 2

= -28x - 20 + 2

= -28x - 18

for

f(x) = 4x-3

g(x) = x2-2

Find f(3a-4)

I will assume g(x) = x^2 - 2

then f(3a-4)

= 4(3a-4) - 3

= 12a - 16 - 3

= 12a - 19

the x in f(x) was merely replaced by 3a-4, so you have to replace it everywhere that you see an x.

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