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

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Answer 1 · 2016-09-19T02:30:18

$f (g (x)) = \sqrt{x - x^{2} + 42}$ $f(g(x)) = sqrt(x-x^2 + 42)$ with domain $- 6 \leq x \leq 7$ $-6 ≤ x ≤ 7$ .

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

$f (g (x)) = \sqrt{42 - (x^{2} - x)}$ $f(g(x)) = sqrt(42 - (x^2 - x))$

$f (g (x)) = \sqrt{- x^{2} + x + 42}$ $f(g(x)) = sqrt(-x^2 +x+ 42)$

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 $0$ $0$ .

$f (g (x)) = \sqrt{- x^{2} - 6 x + 7 x + 42}$ $f(g(x)) = sqrt(-x^2-6x+7x+42 )$

$f (g (x)) = \sqrt{- x (x + 6) + 7 (x + 6)}$ $f(g(x)) = sqrt(-x(x + 6)+ 7(x + 6))$

$f (g (x)) = \sqrt{(7 - x) (x + 6)}$ $f(g(x)) = sqrt((7- x)(x + 6))$

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

$(7 - x) (x + 6) \geq 0$ $(7 - x)(x + 6) ≥ 0$

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 $- 6 \leq x \leq 7$ $-6 ≤ x ≤ 7$ .

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

$7 - x \geq 0$ $7-x>=0$
$x \leq 7$ $x<=7$

$x + 6 \geq 0$ $x+6>=0$
$x \geq - 6$ $x>=-6$

Combining these two, we get $- 6 \leq x \leq 7$ $-6 ≤ x ≤ 7$ .

Hopefully, this helps!

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

Given f(x) = sqrt(42-x) f(x)=√42−xf(x) = sqrt(42-x) and g(x) = x^2 - xg(x)=x2−xg(x) = x^2 - x how do you find f(g(x)) and what is it's domain?