What is the inverse of 4x^2+5,x>= 0 ?

The inverse of 4x^2+5,x>= 0 is (sqrt(x-5))/2 ,-(sqrt(x-5))/2

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inverse of 4x^2+5,x>= 0

inverse

4 x^{2} + 5, x \geq 0

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

Solution steps

y = 4 x^{2} + 5

Replace

x

with

y

x = 4 y^{2} + 5

Solve for

y, x = 4 y^{2} + 5

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

\frac{x}{5 x - 4}

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

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

f (x) = 5 lo g_{e} (x - 10)

\frac{2 + ( x - 3 ) ^{2}}{( x - 3 ) ^{2}}

What is the inverse of 4x^2+5,x>= 0 ?
The inverse of 4x^2+5,x>= 0 is (sqrt(x-5))/2 ,-(sqrt(x-5))/2