Question

Which equation is the inverse of y = 16x^2 + 1?

Answer 1

Switch y and x: x =16y^2 + 1
Solve for y: x-1=16y^2
16y =+-√(x-1)
y =+-√(x-1)/16

Answer 2

Answer:

The inverse of the equation is $f^{-1}(x)=\pm\sqrt{\frac{x-1}{16}}$                

Step-by-step explanation:

Given : Equation $y=16x^2+1$

To find : The inverse of the equation ?

Solution :

Equation $y=16x^2+1$

To find the inverse we interchange the value of x and y,

$x=16y^2+1$

Now, solve for y

$16y^2=x-1$

Divide by 16 both side,

$y^2=\frac{x-1}{16}$

Taking root both side,

$y=\pm\sqrt{\frac{x-1}{16}}$

$f^{-1}(x)=\pm\sqrt{\frac{x-1}{16}}$

Therefore, The inverse of the equation is $f^{-1}(x)=\pm\sqrt{\frac{x-1}{16}}$