What is the inverse of f(x)= 1/16 (x-1)^2-4 ?

The inverse of f(x)= 1/16 (x-1)^2-4 is 4sqrt(x+4)+1,-4sqrt(x+4)+1

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inverse of f(x)= 1/16 (x-1)^2-4

inverse

f (x) = \frac{1}{16} (x - 1)^{2} - 4

4 x + 4 + 1, - 4 x + 4 + 1

Solution steps

y = \frac{1}{16} (x - 1)^{2} - 4

Replace

x

with

y

x = \frac{1}{16} (y - 1)^{2} - 4

Solve for

y, x = \frac{1}{16} (y - 1)^{2} - 4

4 x + 4 + 1, - 4 x + 4 + 1

f (x) = \frac{1 - e ^{x}}{1 + e ^{x}}

f (x) = \frac{13}{x - 7}

f (x) = 12 - x^{2}

2 x^{3} - 6 x^{2} + 4

f (z) = \frac{46}{x}

What is the inverse of f(x)= 1/16 (x-1)^2-4 ?
The inverse of f(x)= 1/16 (x-1)^2-4 is 4sqrt(x+4)+1,-4sqrt(x+4)+1