Question

g(x) = 5x^2 - 4x
h(x) = 3x + 9
Find: g(h(x))

Answer 1

\[\large g(\color{royalblue}{x})=5\color{royalblue}{x}^2-4\color{royalblue}{x}\]

\[\large \color{orangered}{f(x)=3x+9}\]

So let's take f(x) and stuff it inside of our function g,
\[\large g(\color{orangered}{f(x)})=5\color{orangered}{f(x)}^2-4\color{orangered}{f(x)}\]
Plugging in our f(x) gives us,
\[\large g(\color{orangered}{3x+9})=5\color{orangered}{(3x+9)}^2-4\color{orangered}{(3x+9)}\]

Answer 2

Woops, I called it f(x) instead of h(x). No big deal though.
Lemme know if the concept is confusing.

Answer 3

i think i got it thanks (:

Answer 4

wait wat do we plug in for h?

Answer 5

@zepdrix

Answer 6

we plug h in for h c:
h(x) is 3x+9.

So instead of plugging x into g( )
we plug h(x) into it.
Or in other words we plug (3x+9) into g( )

g(3x+9)

everywhere you see an x in your `g` function, replace it with 3x+9

Answer 7

Then you have some work to do simplifying it.

Answer 8

\[\large g(\color{royalblue}{x})=5\color{royalblue}{x}^2-4\color{royalblue}{x}\]

Instead of evaluating the function at `x`, as we've done above, we want to evaluate it at h(x). So we plug h(x) into g(x).

\[\large g(\color{royalblue}{3x+9})=5\color{royalblue}{(3x+9)}^2-4\color{royalblue}{(3x+9)}\]

See how the blue term changed?
Is this a bit confusing? :C Function notation takes a bit of getting used to.