Question

f(x) = Square root of quantity x plus seven.; g(x) = 8x - 11
Find f(g(x)).
A. f(g(x)) = 2Square root of quantity two x plus one.
B. f(g(x)) = 8Square root of quantity x plus seven. - 11
C. f(g(x)) = 8Square root of quantity x minus four.
D. f(g(x)) = 2Square root of quantity two x minus one.

Answer 1

@jim_thompson5910

Answer 2

This is f(x) right?
\[\Large f(x) = \sqrt{x+7}\]
The 7 is under the square root?

Answer 3

correct :)

Answer 4

Replace every copy of 'x' with g(x). Then replace the g(x) on the right side with 8x-11
\[\Large f(x) = \sqrt{x+7}\]
\[\Large f({\color{red}{x}}) = \sqrt{{\color{red}{x}}+7}\]
\[\Large f({\color{red}{g(x)}}) = \sqrt{{\color{red}{g(x)}}+7}\]
\[\Large f({\color{red}{g(x)}}) = \sqrt{{\color{red}{8x-11}}+7}\]
I'll let you simplify

Answer 5

would the square root be.. 8x minus 4? so C?

Answer 6

what does 8x-11+7 simplify to?

Answer 7

I keep getting 8x-4. I did -11+7 and got -4

Answer 8

\[\Large f(g(x)) = \sqrt{8x-11+7}\]
\[\Large f(g(x)) = \sqrt{8x-4}\]
\[\Large f(g(x)) = \sqrt{4(2x-1)}\]
\[\Large f(g(x)) = \sqrt{4}*\sqrt{2x-1}\]
I'll let you do the last step

Answer 9

-8?

Answer 10

what is the square root of 4?

Answer 11

2

Answer 12

so
\[\Large f(g(x)) = \sqrt{4}*\sqrt{2x-1}\]
\[\Large f(g(x)) = 2\sqrt{2x-1}\]

Answer 13

so D?

Answer 14

correct

Answer 15

awesome sauce! thanks!

Answer 16

sure thing