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

2 Answers
F. Javier B.
Jun 1, 2018

See below for explanation

Explanation:

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

We note that #s(x)=x^2+5x#

We proceed as follows

#s(t(x))=s(sqrt(x+4))=(sqrt(x+4))^2+5sqrt(x+4)=x+4+5sqrt(x+4)#

Then #s(t(6))=6+4+5sqrt(6+4)=10+5sqrt10#

A
Jun 1, 2018

#s(t(6))= 10+5sqrt(10) #
This question asks for s(t(6)) not s(t)(6)

Explanation:

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

Additional info:
Note that functions like s(t(x)) are called composite functions.

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