What happens to the slope when the line gets close

Ответ:

Walkman2092

18.04.2020

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

Step-by-step explanation:

The given equation is

y=16x^2+1

For this function to have an inverse, we must restrict the domain, say $x\ge0$

We interchange x and y to get,

x=16y^2+1

We now make y the subject to get;

x-1=16y^2

We divide through by 16 to get;

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

We now take the square root of both sides to get;

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

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

Since $x\ge 0$ , the inverse function is

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

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