Question

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

Answer 1

$f(x)=-4\sqrt{x}-1;\ x\geq0\\\\y=-4\sqrt{x}-1\\\\\text{solve for x}\\\\-4\sqrt{x}-1=y\ \ \ \ |+1\\\\-4\sqrt{x}=y+1\ \ \ \ |:(-4)}\\\\\sqrt{x}=-\dfrac{y+1}{4}\to y+1\leq0\to y\leq-1\\\\\text{square both sides}\\\\x=\dfrac{(y+1)^2}{4^2}\\\\x=\dfrac{(y+1)^2}{16}$

$f^{-1}(x)=\dfrac{(x+1)^2}{16};\ x\leq-1$

Answer 2

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) = $\frac{(x + 1)^{2} }{16}$  ; x ≤ -1