Question

If and g(x)=3, then g[f(x)] =

3x^3-9x^2-3x+15

3x^2-9x+1

-6

1

3

Answer 1

If and g(x)=3, then g[f(x)] =
^
\(\huge ?\)

Answer 2

f(x)=x^2-3x^2-2x+5

Answer 3

if say
g(x) = 2x and f(x) = 5

what would you get for g( f(x) ) ?

Answer 4

10

Answer 5

how did you get that?

Answer 6

plug f into x

Answer 7

\(\bf g(x)=2x\qquad {\color{red}{ f(x)}}=5\\ \quad \\
g(\quad {\color{red}{ f(x)}}\quad )\implies g({\color{red}{ 5}})=2({\color{red}{ 5}})=10\)

Answer 8

yea

Answer 9

you're correct, notice that, we need a variable to "replace", for our value

so if \(\bf g(x)=3\qquad f(x)=x^2-3x^2-2x+5\\ \quad \\
g(\quad f(x)\quad )\implies g(\quad x^2-3x^2-2x+5\quad )\)

but g(x) is just a number, it has no variables, or "placeholders" to replace our f(x) with
so, it remains "3" :)

Answer 10

relly? i thought there was a catch

Answer 11

well, a red herring really
you'd think g(x) will expand to something
but that will only occur if there were variables or "placeholders" to stick the argument, that is f(x), into

since g(x) is just a number, it has no variables, thus no placeholders, f(x) goes nowhere

Answer 12

ok thanks

Answer 13

yw