What is the inverse of f(x)=2x^2+1,x<= 0 ?

The inverse of f(x)=2x^2+1,x<= 0 is sqrt((x-1)/2),-sqrt((x-1)/2)

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inverse of f(x)=2x^2+1,x<= 0

inverse

f (x) = 2 x^{2} + 1, x \leq 0

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

Solution steps

y = 2 x^{2} + 1

Replace

x

with

y

x = 2 y^{2} + 1

Solve for

y, x = 2 y^{2} + 1

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

f (x) = e^{5 x + 1}

f (x) = \frac{2 x + 3}{4 x + 5}, x \neq = - \frac{5}{4}

f (x) = \frac{( x + 2 )}{5 x - 1}

f (x) = 4^{2 x + 1}

f (x) = 2 x - 1 - 8

What is the inverse of f(x)=2x^2+1,x<= 0 ?
The inverse of f(x)=2x^2+1,x<= 0 is sqrt((x-1)/2),-sqrt((x-1)/2)