MATH SOLVE

2 months ago

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

Accepted Solution

A:

Answer:[tex]y=\frac{\sqrt{x-1} }{4}[/tex]Step-by-step explanation:The given equation is [tex]y=16x^2+1[/tex]For this function to have an inverse, we must restrict the domain, say [tex]x\ge0[/tex]We interchange x and y to get,[tex]x=16y^2+1[/tex]

We now make y the subject to get;

[tex]x-1=16y^2[/tex]

[tex]x-1=16y^2[/tex]We divide through by 16 to get;

[tex]\frac{x-1}{16}=y^2[/tex]

We now take the square root of both sides to get;[tex]\pm \sqrt{\frac{x-1}{16}}=y[/tex]

[tex]y=\pm \frac{\sqrt{x-1} }{4}[/tex]

Since Β [tex]x\ge 0[/tex], the inverse function is

[tex]y=\frac{\sqrt{x-1} }{4}[/tex]

We now make y the subject to get;

[tex]x-1=16y^2[/tex]

[tex]x-1=16y^2[/tex]We divide through by 16 to get;

[tex]\frac{x-1}{16}=y^2[/tex]

We now take the square root of both sides to get;[tex]\pm \sqrt{\frac{x-1}{16}}=y[/tex]

[tex]y=\pm \frac{\sqrt{x-1} }{4}[/tex]

Since Β [tex]x\ge 0[/tex], the inverse function is

[tex]y=\frac{\sqrt{x-1} }{4}[/tex]