What is the inverse of f(x)=sqrt(2x^2+5) ?

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

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inverse of f(x)=sqrt(2x^2+5)

inverse

f (x) = 2 x^{2} + 5

\frac{x ^{2} - 5}{2}, - \frac{x ^{2} - 5}{2}

Solution steps

y = 2 x^{2} + 5

Replace

x

with

y

x = 2 y^{2} + 5

Solve for

y, x = 2 y^{2} + 5

\frac{x ^{2} - 5}{2}, - \frac{x ^{2} - 5}{2}

f (x) = 5 x^{3} - 8

f (x) = - x^{2} + 8 x + 41

f (x) = \frac{3}{\frac{x}{x + 3}}

\frac{( x - 2 ) ^{2}}{x - 2}

\frac{2 x - 1}{3 x - 5}

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