Pre calc please help I’m failing this class !!

[tex]f(x)=-4\sqrt{x}-1;\ x\geq0\\\\y=-4\sqrt{x}-1\\\\\text{solve for x}\\\\-4\sqrt{x}-1=y\ \ \ \ |+1\\\\-4\sqrt{x}=y+1\ \ \ \ |:(-4)}\\\\\sqrt{x}=-\dfrac{y+1}{4}\to y+1\leq0\to y\leq-1\\\\\text{square both sides}\\\\x=\dfrac{(y+1)^2}{4^2}\\\\x=\dfrac{(y+1)^2}{16}[/tex]
[tex]f^{-1}(x)=\dfrac{(x+1)^2}{16};\ x\leq-1[/tex]
Inverse is where you swap the x's with y's and the y with x. Then solve for "y". Remember that f(x) is actually "y"
y = -4 √x - 1 → x = -4√y - 1 notice that y ≥ 0
x + 1 = -4√y
- (x + 1)/4 = √y
[- (x + 1)/4]² = (√y)²
(x + 1)²/16 = |y|
(x + 1)²/16 = +/- y
(x + 1)²/16 is positive so disregard "-y"
Now, let's look at the restrictions: x = -4√y - 1 since y≥0, then y must be zero or positive so √y must be zero or positive, therefore -4√y must be zero or negative.
x + 1 = -4√y → x + 1 = 0 or x + 1 = negative
x = -1 or x = negative - 1
Thus, x ≤ -1
Answer: f⁻¹(x) = [tex]\frac{(x + 1)^{2} }{16}[/tex] ; x ≤ -1