Question

Find all real numbers (x) (if any) such that F(g(x)) = g(f(x)). Where F(x) = x^2+1 AND g(X) = 1 - X

Answer 1

Find f(g(x)) and g(f(x)) first.

Equate them and then solve for x.

Answer 2

Okay I put \[x ^{2} + 1 = 1 - x\] . Then, Added the single X. and Subtracted the single (+) 1. to end up at \[x ^{3} = 0\]. So... For the function of F = All real numbers. and for G = All real numbers above (greater than) 1?

Answer 3

When you're adding an X-> You do X^{2} right? not 2x. .... Ummm The part where I added the single X. Did I do that right or did i do that wrong?

Answer 4

Need clarification on something. Does the problem have uppercase and lowercase F and f? Because then it will be two different functions.

Answer 5

Lower case

Answer 6

f(x) = x^2 + 1
g(x) = 1 - x
f(g(x)) = (1-x)^2 + 1 = 1 + x^2 - 2x + 1 = x^2 - 2x + 2
g(f(x)) = 1 - (x^2 + 1) = 1 - x^2 - 1 = -x^2
f(g(x)) = g(f(x))
x^2 - 2x + 2 = -x^2
2x^2 - 2x + 2 = 0
x^2 - x + 1 = 0
solve for x.

Answer 7

f(x) = x^2 + 1
To find f(g(x)), replace x with g(x):
f(g(x)) = (g(x))^2 + 1 = (1-x)^2 + 1 (as done above)

Answer 8

So, do i have to factor at \[x ^{2}-x+1 = 0\]

Answer 9

Use the quadratic formula.

Answer 10

If you find the discriminant, b^2 - 4ac, you will notice it is negative. Therefore, there are no real values of x that will satisfy the condition.

Answer 11

So since it cannot be factored, there is no real number that can satisfy X. So the answer is No real numbers?

Answer 12

The answer is "No REAL Numbers" but not because it cannot be factored. When we use the quadratic formula we get a negative number inside the square root. That means the equation has complex roots and no real roots.

Answer 13

Thank You for your help :)

Answer 14

You are welcome.