What is the inverse of f(x)=5y+4=(x+3)^2+1/2 ?

The inverse of f(x)=5y+4=(x+3)^2+1/2 is sqrt((2x-1)/2)-3,-sqrt((2x-1)/2)-3

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inverse of f(x)=5y+4=(x+3)^2+1/2

inverse

f (x) = 5 y + 4 = (x + 3)^{2} + \frac{1}{2}

\frac{2 x - 1}{2} - 3, - \frac{2 x - 1}{2} - 3

Solution steps

y = (x + 3)^{2} + \frac{1}{2}

Replace

x

with

y

x = (y + 3)^{2} + \frac{1}{2}

Solve for

y, x = (y + 3)^{2} + \frac{1}{2}

\frac{2 x - 1}{2} - 3, - \frac{2 x - 1}{2} - 3

\frac{36 - x ^{2}}{x + 2}

f (x) = 3 x - 4

f (x) = \frac{2 x - 4}{3}

\frac{7}{x}

(0, 2), (1, 1)

What is the inverse of f(x)=5y+4=(x+3)^2+1/2 ?
The inverse of f(x)=5y+4=(x+3)^2+1/2 is sqrt((2x-1)/2)-3,-sqrt((2x-1)/2)-3