Question

Which equation is the inverse of y = 16x2 + 1?

Answer 1

we have

$y=16x^{2} +1$

Step 1

exchange variables x for y and y for x

$x=16y^{2} +1$

Step 2

Isolate the variable y

Subtract $1$ both sides

$x-1=16y^{2}+1-1$

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

Divide by $16$ both sides

$16y^{2}/16=(x-1)/16$

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

Square root both sides

$y=(+/-)\sqrt{\frac{x-1}{16}}$

$y=(+/-)\frac{\sqrt{x-1}}{4}$

Step 3

Let

$f(x)^{-1}=y$

$f(x)^{-1}=(+/-)\frac{\sqrt{x-1}}{4}$

therefore

the answer is

The inverse is equal to

$f(x)^{-1}=(+/-)\frac{\sqrt{x-1}}{4}$

Answer 2

y = 16x^2 + 1,  We can divide both site of the equation to find the value of x.  You can do any valid operation on both sides of the equation and it will still be equal. 

y - 1 = 16x^2     Divide both side by 16. 

$\frac{ \sqrt{y - 1}}{4}$ = x

The inverse is 
$f^{-1}$ (x) = $\frac{ \sqrt{x - 1}}{4}$