let f(x)=x^2-2x and g(x)=square root of (1-x). what is f(g(x)) and what is the domain of f(g(x))?

Question

Please please help with this, I am really confused... thanks!

Answer 1

Hi!

to understand this problem it is important to know that f(g(x)) means that every x in the f(x) expression is replaced by the expression of g(x)

f(x) = x²- 2x and g(x)= √(1-x)

f(g(x)) = ( √(1-x) )² - 2(√(1-x) )

0 = 1-x - (2 (√(1-x) )

subtract 1-x from both sides

x-1 = - (2 (√(1-x) )

Divide -2 by both sides

( (x-1) / -2 ) = √(1-x)

To remove square root take the square of both sides

( (x-1) / -2 )² = 1-x

(x-1)² / (-2)² = 1-x

(x-1) (x-1) / (-2)² = 1-x

(x² - 2x + 1) / 4 = 1-x

Multiply by 4 on both sides

(x² - 2x + 1) = 4 - 4x

x² + 2x - 3 = 0

Factor

(x-1)(x+3) =0

zeros = 1 and -3

x intercepts (1,0) and (-3,0)

the domain will be x≤1 or (-∞,1)

domain shortcut: if you graph this function (desmos works great) it is easy to find the domain

Answer 2

We have two functions:

f(x) = x^2 - 2x

g(x) = sqrt(1-x)

f(x) is defined everywhere.

g(x) is defined when x >= 1, since if x < 1, you would be taking the square root of a negative number (don't do that!)

However, we're looking at f(g(x))--so we want to look at f(x) = x^2 - 2x, and plug in for x this entire function we called g. Wherever you see an x, replace it with sqrt(1-x).

So we literally want to do (sqrt(1-x))^2 - 2*sqrt(1-x). Let's simplify that.

sqrt(1-x)^2 is just (1-x), so it's (1-x) - 2*sqrt(1-x).

In other words, f(g(x)) = -x - 2*sqrt(1-x) + 1

Again, this is defined for x >= 1