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

2 Answers
Alan P.
Nov 27, 2015

#sot(6) = 10+5sqrt(10) = 5(2+sqrt(10))#

Explanation:

Since #s(color(blue)(x)) = color(blue)(x)^2+2color(blue)(x)+3color(blue)(x) = color(blue)(x)^2+5color(blue)(x)#

#sot(6)# (also written #s(color(blue)(t(6)))#)
#color(white)("XXX")=color(blue)(t(6))^2+5*color(blue)(t(6))#

And since #t(color(red)(x)) =sqrt(color(red)(x)+4)#
#color(white)("XXX")t(color(red)(6)) = sqrt(color(red)(6)+4) = sqrt(10)#

Therefore:
#color(white)("XXX")sot(6)=s(t(6)) = (sqrt(10))^2+5*sqrt(10) = 10+5sqrt(10)#

Tony B
Nov 27, 2015

Assuming you meant: #s(x)=x^2+2x+3 " and "t(x)=sqrt(x+4)#
Find #s@t(6)#

#color(blue)(s@t(6)=13+2sqrt(10))#

Explanation:

Assumption: There is a typing error in the question and that it should be:

#s(x)=x^2+2x+3# and #t(x)=sqrt(x+4)#

Find #s@t(6)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)(Step 1)#

#t(x) -> t(6) = sqrt(6+4) =sqrt(10)#

#color(blue)(Step2)#

#s(x)->s@t(6) =(sqrt(10))^2+2(sqrt(10))+3#

#s@t(6)=13+2sqrt(10)#

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