What is the inverse of f(x)=16(3-x)^2-1 ?

The inverse of f(x)=16(3-x)^2-1 is -(sqrt(x+1)-12)/4 ,-(-sqrt(x+1)-12)/4

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inverse of f(x)=16(3-x)^2-1

inverse

f (x) = 16 (3 - x)^{2} - 1

- \frac{x + 1 - 12}{4}, - \frac{- x + 1 - 12}{4}

Solution steps

y = 16 (3 - x)^{2} - 1

Replace

x

with

y

x = 16 (3 - y)^{2} - 1

Solve for

y, x = 16 (3 - y)^{2} - 1

- \frac{x + 1 - 12}{4}, - \frac{- x + 1 - 12}{4}

What is the inverse of f(x)=16(3-x)^2-1 ?
The inverse of f(x)=16(3-x)^2-1 is -(sqrt(x+1)-12)/4 ,-(-sqrt(x+1)-12)/4