Answer:
![y=\frac{\sqrt{x-1} }{4}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%5Csqrt%7Bx-1%7D%20%7D%7B4%7D)
Step-by-step explanation:
The given equation is
![y=16x^2+1](https://tex.z-dn.net/?f=y%3D16x%5E2%2B1)
For this function to have an inverse, we must restrict the domain, say ![x\ge0](https://tex.z-dn.net/?f=x%5Cge0)
We interchange x and y to get,
![x=16y^2+1](https://tex.z-dn.net/?f=x%3D16y%5E2%2B1)
We now make y the subject to get;
![x-1=16y^2](https://tex.z-dn.net/?f=x-1%3D16y%5E2)
![x-1=16y^2](https://tex.z-dn.net/?f=x-1%3D16y%5E2)
We divide through by 16 to get;
![\frac{x-1}{16}=y^2](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7B16%7D%3Dy%5E2)
We now take the square root of both sides to get;
![\pm \sqrt{\frac{x-1}{16}}=y](https://tex.z-dn.net/?f=%5Cpm%20%5Csqrt%7B%5Cfrac%7Bx-1%7D%7B16%7D%7D%3Dy)
![y=\pm \frac{\sqrt{x-1} }{4}](https://tex.z-dn.net/?f=y%3D%5Cpm%20%5Cfrac%7B%5Csqrt%7Bx-1%7D%20%7D%7B4%7D)
Since
, the inverse function is
![y=\frac{\sqrt{x-1} }{4}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%5Csqrt%7Bx-1%7D%20%7D%7B4%7D)