# Let s(x) = x^2 + 2x + 3x and t(x) =sqrt( x+4) how do you find (s( t)) (6)?

Jun 1, 2018

See below for explanation

#### Explanation:

We have two functions and we have to compose them in the order to find $s \left(t \left(6\right)\right)$. (The composition of functions is not conmutative)

We note that $s \left(x\right) = {x}^{2} + 5 x$

We proceed as follows

$s \left(t \left(x\right)\right) = s \left(\sqrt{x + 4}\right) = {\left(\sqrt{x + 4}\right)}^{2} + 5 \sqrt{x + 4} = x + 4 + 5 \sqrt{x + 4}$

Then $s \left(t \left(6\right)\right) = 6 + 4 + 5 \sqrt{6 + 4} = 10 + 5 \sqrt{10}$

Jun 1, 2018

$s \left(t \left(6\right)\right) = 10 + 5 \sqrt{10}$
This question asks for s(t(6)) not s(t)(6)

#### Explanation:

To solve for $s \left(t\right) \left(6\right)$
$s \left(x\right) = {x}^{2} + 2 x + 3 \cdot x$
$t \left(x\right) = \sqrt{x + 4}$
let's first find s(t).
Substitute t(x) into s(x):
$s \left(t \left(x\right)\right) = {\left(\sqrt{x + 4}\right)}^{2} + 2 \sqrt{x + 4} + 3 \cdot \sqrt{x + 4}$
Simplifying this gives us:
$s \left(t \left(x\right)\right) = \left(x + 4\right) + 2 \sqrt{x + 4} + 3 \cdot \sqrt{x + 4}$
Now, subbing 6 into the X values gives us:
$s \left(t \left(6\right)\right) = \left(6 + 4\right) + 2 \sqrt{6 + 4} + 3 \cdot \sqrt{6 + 4}$
$s \left(t \left(6\right)\right) = 10 + 2 \sqrt{10} + 3 \cdot \sqrt{10}$
$s \left(t \left(6\right)\right) = 10 + 5 \sqrt{10}$