f(x)=2+4x-x^2 when x is less than 2
thus f-1(x)=2-sqrt(-x+6)
i need to show that f(f-1(x))=x, but i keep getting negative x...any help
by definition ...
f(f^-1 (x)) = x , but anyway ...
f( f^-1(x) )
= f(2 - √(6-x) )
= 2 + 4(2 - √(6-x) ) - (2 - √(6-x) )^2
= 2 + 8 - 4√(6-x) - (4 - 4√(+-x) + 6-x)
= 10 - 4√(6-x) - 4 + 4√(6-x) - 6 + x
= x
To show that f(f-1(x)) = x, we need to first find f-1(x) and then substitute it back into f(x).
Given f(x) = 2 + 4x - x^2 when x is less than 2, we can find the inverse function f-1(x).
To find the inverse, we replace f(x) with y and solve for x:
y = 2 + 4x - x^2
Rearrange the equation to isolate x:
0 = x^2 - 4x + 2 - y
Now, let's use the quadratic formula to solve for x:
x = (-(-4) ± √((-4)^2 - 4(1)(2 - y))) / (2(1))
Simplifying:
x = (4 ± √(16 - 4(2 - y))) / 2
x = (4 ± √(16 - 8 + 4y)) / 2
x = (4 ± √(8 + 4y)) / 2
x = (2 ± √(2 + y))
Since f-1(x) represents the inverse function, we choose the minus sign:
f-1(x) = 2 - √(2 + x)
Note: We don't include the case with the plus sign because it is outside the given domain of x less than 2.
Now, substitute f-1(x) back into f(x):
f(f-1(x)) = 2 + 4(2 - √(2 + x)) - (2 - √(2 + x))^2
Simplify the expression:
f(f-1(x)) = 2 + 8 - 4√(2 + x) - 2(√(2 + x)) + (2 + x)
Combine like terms:
f(f-1(x)) = 12 - 6√(2 + x) + x
Now, let's check if f(f-1(x)) equals x:
12 - 6√(2 + x) + x = x
Rearrange the equation:
12 - 6√(2 + x) = 0
6√(2 + x) = 12
Divide by 6:
√(2 + x) = 2
Square both sides:
2 + x = 4
x = 2
Therefore, f(f-1(x)) = x holds when x equals 2.