What is the inverse of g(x)=2(x-4)^2+1 ?

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

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inverse of g(x)=2(x-4)^2+1

inverse

g (x) = 2 (x - 4)^{2} + 1

\frac{x - 1}{2} + 4, - \frac{x - 1}{2} + 4

Solution steps

y = 2 (x - 4)^{2} + 1

Replace

x

with

y

x = 2 (y - 4)^{2} + 1

Solve for

y, x = 2 (y - 4)^{2} + 1

\frac{x - 1}{2} + 4, - \frac{x - 1}{2} + 4

y = 4 x^{3} - 3

y = 0.5 x - 1.5

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

y = \frac{- 2 x + 2}{x + 7}

\frac{- 1}{\frac{1}{6} \cdot 8 3}

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