What is the inverse of f(x)=(sqrt(x+3))/4-5 ?

The inverse of f(x)=(sqrt(x+3))/4-5 is 16x^2+160x+397

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inverse of f(x)=(sqrt(x+3))/4-5

inverse

f (x) = \frac{x + 3}{4} - 5

16 x^{2} + 160 x + 397

Solution steps

y = \frac{x + 3}{4} - 5

Replace

x

with

y

x = \frac{y + 3}{4} - 5

Solve for

y, x = \frac{y + 3}{4} - 5

16 x^{2} + 160 x + 397

f (x) = 2 (x + 4)^{3}

\frac{x + 2}{x - 1}

f (x) = \frac{2}{x - 3}

2 + x

What is the inverse of f(x)=(sqrt(x+3))/4-5 ?
The inverse of f(x)=(sqrt(x+3))/4-5 is 16x^2+160x+397