Question

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

Answer 1

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

`f^(-1)(x)=((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) =
`((x + 1)^(2) )/(16)` ; x ≤ -1