What is the inverse of f(x)=x^2-5,x>= 0 ?

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

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inverse of f(x)=x^2-5,x>= 0

inverse

f (x) = x^{2} - 5, x \geq 0

x + 5, - x + 5

Solution steps

y = x^{2} - 5

Replace

x

with

y

x = y^{2} - 5

Solve for

y, x = y^{2} - 5

x + 5, - x + 5

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

5 p + 2 q = 4

\frac{x - 6}{x ^{2} - 36}

(48, 100), (42, 125)

\frac{x - 5}{x ^{2} + 25} - 3 x

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