What is the inverse of h(x)=x^2-4,x>= 0 ?

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

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inverse of h(x)=x^2-4,x>= 0

inverse

h (x) = x^{2} - 4, x \geq 0

x + 4, - x + 4

Solution steps

y = x^{2} - 4

Replace

x

with

y

x = y^{2} - 4

Solve for

y, x = y^{2} - 4

x + 4, - x + 4

\frac{2 x - 9}{x - 4}

\frac{2 x + 1}{2 x}

f (x) = - 0.8 x + 188

f (x) = 1.7

f (x) = 4.5 - 2 a^{0.5} (a - x)^{0.5}

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