Having trouble with this question on my assignment. I got the rest answered.
Assume a domain A = {x ∈ ℝ | x ≥ 2} for the function f(x) = x2 − 4x + 2.
Find the value of the inverse f -1(x) at x = 5. Provide 4 decimal places.
I'm getting 5.0000 but I'm sure that it's wrong.
Corrections:
f(x) = x^2 − 4x + 2
f^-1(x) at x = 5.
well, is f(5) = 5? No.
So, f^-1(5) is not 5.
y = x^2-4x+2 = (x-2)^2 - 2
The inverse function, restricted to the domain above, is
f^-1(x) = 2+√(x+2)
so, f^-1(5) = 2+√7
To find the value of the inverse function f^(-1)(x) at x = 5, we first need to find the inverse function f^(-1)(x).
To do this, we will follow these steps:
Step 1: Replace f(x) with y in the original equation:
y = x^2 - 4x + 2
Step 2: Swap x and y:
x = y^2 - 4y + 2
Step 3: Solve for y:
x = y^2 - 4y + 2
Rearrange the equation:
y^2 - 4y + (2 - x) = 0
Step 4: Solve the quadratic equation for y using the quadratic formula:
y = [4 ± √(4^2 - 4(1)(2-x))]/(2)
Simplifying further, we have:
y = [4 ± √(16 - 4(2 - x))]/2
y = [4 ± √(16 - 8 + 4x)]/2
y = [4 ± √(4x + 8)]/2
y = (4 ± √(4x + 8))/2
y = 2 ± √(x + 2)
Now that we have the inverse function f^(-1)(x) as y = 2 ± √(x + 2), we can substitute x = 5 to find the value of the inverse at that point.
When x = 5:
f^(-1)(5) = 2 ± √(5 + 2)
f^(-1)(5) = 2 ± √(7)
Since we need the value with 4 decimal places, we can approximate the square root of 7 as 2.6458.
Therefore, f^(-1)(5) = 2 ± 2.6458 = 4.6458 or -0.6458.
So, the value of the inverse function f^(-1)(x) at x = 5 is approximately 4.6458 or -0.6458.