Given #f(x) = sqrt(42-x) # and #g(x) = x^2 - x# how do you find f(g(x)) and what is it's domain?

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Answer 1 · 2016-09-19T02:30:18

#f(g(x)) = sqrt(x-x^2 + 42)# with domain #-6 ≤ x ≤ 7#.

First, find the composition by inputting #g(x)# inside #f(x)#.

#f(g(x)) = sqrt(42 - (x^2 - x))#

#f(g(x)) = sqrt(-x^2 +x+ 42)#

To determine the domain, we need to factor inside the square root, because we cannot have the value inside the square root be any less than #0#.

#f(g(x)) = sqrt(-x^2-6x+7x+42 )#

#f(g(x)) = sqrt(-x(x + 6)+ 7(x + 6))#

#f(g(x)) = sqrt((7- x)(x + 6))#

Set up a quadratic inequality, forgetting about the square root for the time being.

#(7 - x)(x + 6) ≥ 0#

Select test points. Let Test Point 1 be 9, test point 2 be 5 and test point 3 be -8.

Checking, you will find test point 1 does not work, test point 2 works and test point 3 does not work.

Hence, the domain is #-6 ≤ x ≤ 7#.

Another way to work it out is to solve the inequality one bracket at a time.

#7-x>=0#
#x<=7#

#x+6>=0#
#x>=-6#

Combining these two, we get #-6 ≤ x ≤ 7#.

Hopefully, this helps!

Here's your graph: graph{sqrt(-x^2+x+42) [-9.59, 10.41, -0.36, 9.64]}